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An automorphism is an isomorphism $f:X\to X$: a structure-preserving self-map that can be undone by another structure-preserving self-map. Equivalently, it is an invertible endomorphism. Its inverse satisfieswhere $1_X$ is the identity morphism. Both maps must be morphisms in the chosen category. For ordinary groups, rings, and vector spaces over a fixed field, a bijective morphism automatically has a compatible inverse. For topological spaces, an automorphism is a homeomorphism from the space to itself: a continuous bijection needs a continuous inverse as well. See Riehl, Category Theory in Context, Definition 1.1.10 and Example 1.1.11.

For example, $T(x,y)=(x+y,y)$ is an automorphism of the vector space ${\mathbb{R}}^2$ over the real numbers, with inverse $T^{-1}(u,v)=(u-v,v)$. Both are linear maps. The linear endomorphism $P(x,y)=(x,0)$ from endomorphism preserves addition and scalar multiplication too, but discards a coordinate and cannot be undone. Preserving the structure and preserving enough information for reversal are separate requirements.

The automorphisms of a fixed object in a locally small category form a group under composition in a category: the identity morphism is the identity element, and the inverse element of a composite reverses the order of its factors. This makes reversible internal changes calculable. For a group, the resulting automorphism group records which relabelings preserve all products; the examples in group automorphism show how restrictive that condition can be.

Write the finite cyclic group as $C_n=⟨r⟩$. A group homomorphism is determined by $r\mapsto r^a$; it is a group automorphism exactly when $r^a$ generates $C_n$, equivalently $\gcd (a,n)=1$. Composition multiplies the exponents modulo $n$, proving $\mathrm{Aut}(C_n)\cong (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Its group order is therefore Euler's totient function $\varphi (n)$. A group homomorphism $C_2\to \mathrm{Aut}(C_n)$ is determined by a unit $a$ with $a^2\equiv 1(\mathrm{mod}n)$.

The commutator subgroup of a group $G$ is the subgroup generated by all its group commutators:Here and throughout, $[x,y]=x^{-1}{y}^{-1}xy$. A group commutator is the identity element precisely when $x$ and $y$ commute. The generated subgroup includes finite products of group commutators and their inverse elements; an element of $G^{\prime }$ need not itself be a single group commutator. Thus $G^{\prime }=\{e\}$ precisely when $G$ is an abelian group.

The useful idea is to record exactly what must disappear to make multiplication commutative. The identity $xy=yx[x,y]$ shows how a group commutator corrects a change in multiplication order. Setting every such correction equal to $e$ produces the abelianization $G/G^{\prime }$, while preserving every piece of information that a group homomorphism to an abelian group can detect. This connects a local comparison of two operations to a canonical quotient group of the whole group.

The commutator subgroup is a characteristic subgroup, hence a normal subgroup: a group automorphism $\alpha$ sends $[x,y]$ to $[\alpha (x),\alpha (y)]$. For any normal subgroup $N$,Indeed, the group commutator of $xN$ and $yN$ is $[x,y]N$. This makes $G^{\prime }$ the smallest normal subgroup whose quotient group is an abelian group. See Conrad, Subgroup series I, section 4: commutators and abelian quotients.

For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

For a subset $S$ of a group and an element $g$, its conjugate subset isConjugation is a bijection and a group automorphism, so it preserves cardinality and every property expressed using the group operation.

Conjugation by $g$ is the group automorphism $x\mapsto gx{g}^{-1}$. Its orbits are the conjugacy classes.

An endomorphism is a morphism $f:X\to X$: its source and target are the same object, with the same specified structure. For a group it is an operation-preserving group homomorphism back into that group; for a vector space it is a linear endomorphism; for a ring it is a ring endomorphism. A self-map must satisfy the chosen compatibility conditions to qualify. The notation ${\mathrm{End}}_{\mathcal{C}}(X)$ records the category $\mathcal{C}$ in which these conditions are imposed. See Riehl, Category Theory in Context, Definition 1.1.10 and the following paragraph.

An endomorphism can lose information or miss part of its target. On the additive integers, $d(k)=2k$ preserves addition and is injective, but is not surjective: no integer maps to $1$. On the vector space ${\mathbb{R}}^2$ over the real numbers, the linear endomorphism $P(x,y)=(x,0)$ loses the second coordinate, so it is neither injective nor surjective. Neither is an automorphism. The same doubling rule becomes a group isomorphism when its target is the additive even integers, as explained under isomorphism.

The finite cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ offers a small example of information loss that can be drawn completely. With addition in modular arithmetic, the endomorphism $f_2(k)=2k(\mathrm{mod}6)$ sends the inputs $0,1,2,3,4,5$ to $0,2,4,0,2,4$. Its image of a group homomorphism is $\{0,2,4\}$ and its kernel of a group homomorphism is $\{0,3\}$. Each reached output has two inputs; the other three outputs are missed.

For an endomorphism of a group or ring with a finite underlying set, injectivity is equivalent to surjectivity by counting; either condition makes it an automorphism. The additive integers example shows why the finiteness assumption matters. A linear endomorphism of a finite-dimensional vector space also has equivalent injectivity and surjectivity, this time by the rank-nullity theorem even when its underlying set is infinite.

Self-maps can be iterated without changing the object: $f^2=f\circ f$, $f^3=f\circ f\circ f$, and so on. Composition in a category and the identity morphism make the endomorphisms of a fixed object in a locally small category into a monoid. This packages repeated compatible operations, including irreversible ones, into a single structure.

The general language has a documented historical milestone in the 1945 paper of Samuel Eilenberg and Saunders Mac Lane. They defined categories of objects and compatible maps, and distinguished maps admitting inverses within that category. Their examples included groups, vector spaces, and topological spaces. This gave a common framework for compatible self-maps and their invertible cases. See Eilenberg and Mac Lane, General Theory of Natural Equivalences, pp. 234 and 237–239.

Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

A group isomorphism is a bijective group homomorphism $\phi :G\to H$. It matches the entire group operation: $\phi (ab)=\phi (a)\phi (b)$, with each product taken in its own group. Its inverse is automatically a group homomorphism. A group automorphism is the special case $H=G$.^[ref]

The target is part of the claim. The doubling example under isomorphism is a group isomorphism $d:\mathbb{Z}\to 2\mathbb{Z}$, between the additive integers and the additive even integers. Exactly the same rule $d(k)=2k$, declared as $d:\mathbb{Z}\to \mathbb{Z}$, is an injective endomorphism that misses $1$ and every other odd integer, so is not a group automorphism. Changing the declared target changes surjectivity, even though it does not change the formula or its compatibility with addition.

An isomorphism lets a calculation in one model solve the same problem in another. The example under group identifies addition modulo four with function composition of rotations of a square via $k\mapsto r^k$. Through this group isomorphism, adding exponents corresponds exactly to composing motions. More generally, a group isomorphism preserves order of a group element, carries subgroups to subgroups, and carries a generating set of a group to a generating set of a group. Having equally many elements is insufficient: the permutation in group automorphism already fails compatibility on two inputs.

Why does the inverse preserve the group operation? Write $u=\phi (a)$ and $v=\phi (b)$, possible because of surjectivity. ThenThis proves the inverse condition directly. Analogous calculations work for each operation of a ring and for addition and scalar multiplication in a vector space. The argument relies on preserving those operations; it supplies no general continuity claim for the inverse of a continuous bijection between topological spaces.

The holomorph acts on the underlying set of a group by $x\mapsto g\alpha (x)$. Its multiplication is $(g,\alpha )(h,\beta )=(g\alpha (h),\alpha \beta )$. This is exactly the normalizer of the left regular permutation representation: a normalizing permutation fixing the identity is an group automorphism, and every normalizing permutation is a left translation followed by one. The embedded copy of $G$ is normal and the identity stabilizer is its group automorphism group.

For a subgroup $H$ of a group $G$, the normalizer isIt is the stabilizer subgroup of $H$ under the conjugation action on subgroups. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the symmetric group $S_3$, the cyclic subgroup $H=⟨r⟩$ has normalizer $S_3$, although the centralizer of $r$ has only three elements, as illustrated in centralizer and normalizer.

Every element of $H$ preserves $H$ under conjugation, so $H\subseteq N_G(H)$. Moreover, $H$ is a normal subgroup of $N_G(H)$, and this is the largest subgroup of $G$ with that property. Indeed, any subgroup in which $H$ is a normal subgroup consists entirely of elements preserving $H$ under conjugation. In particular, $N_G(H)=G$ exactly when $H$ is a normal subgroup of $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct conjugate subsets are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the subgroup counterpart of counting a conjugacy class with a centralizer: the object being stabilized determines which denominator belongs in the count.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ already gives equality by equal cardinalities. For an infinite subgroup, that inference can fail: conjugation might carry it onto a proper subgroup. The normal subgroup criterion is therefore, with the explicit hypothesis $H\le G$,No finiteness or commutativity assumption is needed. A normal subgroup is preserved as a set; its elements need not lie in the center of a group.

There is a second useful count, distinct from counting conjugate subgroups. The normalizer acts on $H$ itself by group automorphisms, giving the group homomorphismIts group kernel is $C_G(H)$, so the centralizer of a subset is a normal subgroup of $N_G(H)$ and the first isomorphism theorem givesThis quotient group records the different rearrangements of $H$ induced by the surrounding group. It gives a faithful group action on $H$, but need not be the entire automorphism group of $H$. For example, take the ambient group $G=H=⟨(123)⟩$. Since $G$ is an abelian group, $N_G(H)=C_G(H)=G$, so the quotient group is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity group automorphism of $H$. Returning to the ambient symmetric group $S_3$ makes that group automorphism available through a reflection, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two cosets are the three rotations and the three reflections.

The subgroup $H$ itself need not be contained in $C_G(H)$: their intersection is the center of a group $Z(H)$. It is contained precisely when $H$ is an abelian group. Thus $N_G(H)/H$, which exists because $H$ is a normal subgroup of its normalizer, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.

An outer group automorphism is not conjugation by any element of the group. Its nonidentity class belongs to $\mathrm{Out}(G)=\mathrm{Aut}(G)/\mathrm{Inn}(G)$. A permutation-group example is the exceptional automorphism of $S_6$ that swaps its natural and transitive copies of $S_5$; it sends a transposition to a product of three disjoint transpositions and hence cannot be inner.

An group automorphism preserves element orders and sizes of conjugacy classes. A nonidentity involution in $S_n$ has cycle type $2^k{1}^{n-2k}$ and class sizeFor its class to have the same size as the transposition class, it must satisfyFor fixed $k\ge 2$ the left side grows strictly with $n$. When $k=2$, it is $(n-2)(n-3)$, which is two at $n=4$ and at least six at $n=5$, never the required four. For $k=3$ equality first occurs at $n=6$, since $4!=2^2\cdot 3!$, and thereafter the left side is larger. For $k\ge 4$ it is already larger at $n=2k$: the ratio $(2k-2)!/[2^{k-1}k!]$ is $15/4$ at $k=4$, and increases with $k$, its successive ratio being $k(2k-1)/(k+1)>1$. This proves the symmetric-group involution class-size collision result. Therefore, for $n\ne 6$, every group automorphism maps transpositions to transpositions.

Now recover the underlying points from transpositions. Two distinct transpositions fail to commute exactly when their two-element supports meet. A maximal family of pairwise intersecting edges of a complete graph is either a vertex star or a triangle. Indeed, after selecting edges $\{1,2\}$ and $\{1,3\}$, every edge either contains 1 or is $\{2,3\}$; if the latter occurs, all edges lie in that triangle. For $n\ge 5$, the stars have size $n-1\ge 4$ and triangles size three, so stars are recognized purely by the multiplication structure of $S_n$.

An group automorphism must thus permute the stars, giving a permutation $\tau$ of the points. The unique transposition shared by stars $i,j$ is $(ij)$, so its image is $(\tau (i)\tau (j))$. Since transpositions generate $S_n$, the whole group automorphism is conjugation by $\tau$.

At $n=4$, stars and triangles both have three elements, but their generated subgroups distinguish them: the transpositions in a star generate all of $S_4$ (since $(ij)=(ai)(aj)(ai)$ for its center $a$), while a triangle generates the $S_3$ fixing the fourth point. The same reconstruction follows. At $n=3$, conjugation by $S_3$ induces every permutation of its three transpositions (a transposition swaps the other two, and a three-cycle rotates all three), so a transposition-preserving group automorphism is again inner. At $n=2$ the only group automorphism of $C_2$ is the identity, and $n=0,1$ are trivial (if the empty-set degree is included). ThusThe small-degree adjustments are necessary: a cardinality-only star argument would not cover $n=4$.

Use the six-point matching geometry. A duad is an edge of the complete graph on six points, a syntheme is a perfect matching of its three pairs, and a pentad is a one-factorization: five pairwise edge-disjoint synthemes covering all fifteen duads.

First count the pentads without assuming their number. There are fifteen synthemes: the partner of the first point has five choices, the least remaining point then has three possible partners, and the last pair is forced. For a fixed syntheme $M$, eight others share no edge with it: inclusion-exclusion gives $15-3\cdot 3+3\cdot 1-1=8$. Two edge-disjoint synthemes $M,M^{\prime }$ have union a six-cycle. The complement of that cycle is a triangular prism, consisting of two triangles with three corresponding cross edges. It has precisely four perfect matchings: the all-cross matching and three matchings using one cross edge plus one edge from each triangle. The all-cross matching leaves two odd triangles and therefore cannot be completed by two matchings. The other three matchings are pairwise edge-disjoint and partition the prism. Hence $M,M^{\prime }$ extend to exactly one pentad.

Each pentad through $M$ uses four of its eight disjoint partners, and every partner gives exactly one completion. Thus $M$ lies in two pentads. Counting incidences gives $15\cdot 2/5=6$ pentads in total. Distinct pentads cannot share two synthemes because two disjoint synthemes determine their unique completion. Every syntheme therefore identifies a distinct pair of pentads; there are fifteen such pairs, so this incidence map is a bijection.

Permuting the six original points permutes the six pentads, defining a group homomorphism $\Phi :S_6\to \mathrm{Sym}(\text{pentads})$. It is faithful. If an element fixes every pentad, it fixes every syntheme by the pair-incidence bijection. A duad is the unique common edge of two distinct synthemes containing it, so every duad is fixed as well. A permutation fixing all two-element subsets fixes every point, by intersecting $\{i,j\}$ and $\{i,k\}$ for distinct $i,j,k$. The kernel of a group homomorphism is therefore trivial. Both groups have order $6!$, so labeling the pentads produces an group automorphism of $S_6$.

It remains to prove this group automorphism is outer. A transposition, say $(12)$, fixes no pentad. If it fixed one, the unique syntheme in that pentad containing the duad $\{1,2\}$ would be fixed. Every other syntheme $M^{\prime }$ in the pentad contains 1 and 2 in separate edges. Swapping them changes those two edges but leaves the third edge unchanged. Hence $(12)M^{\prime }$ is a distinct syntheme sharing an edge with $M^{\prime }$, impossible for two members of the same pentad. Thus $(12)$ acts without fixed points on the six pentads and becomes a product of three disjoint transpositions.

Inner automorphisms preserve cycle type, whereas $\Phi$ sends a single transposition to a triple transposition. ConsequentlyThis realizes the exceptional class-size coincidence from part (a). It is the full symmetric-group version of the pentad construction of the exceptional alternating-group automorphism, with faithfulness proved directly from incidence rather than borrowed from a simplicity theorem.

Retain the identification with the left regular action. A group group automorphism $\phi$ of $G$ satisfies $\phi L_g{\phi }^{-1}=L_{\phi (g)}$, so it normalizes $L(G)$ and fixes the identity point.

Conversely let a normalizing permutation $n$ fix the identity. Its conjugation on $L(G)$ induces a group group automorphism $\phi$ characterized by $n{L}_g{n}^{-1}=L_{\phi (g)}$. Evaluating at the identity gives $n(g)=\phi (g)$, so the whole identity stabilizer in the normalizer is exactly $\mathrm{Aut}(G)$. For arbitrary $n$, put $t=n(1)$. Then $L_t^{-1}n$ fixes the identity and is an group automorphism; thus every element has a unique expression $L_t\phi$.

The translation subgroup is normal in its normalizer, its intersection with the group automorphism subgroup is trivial, and the multiplication rule isConsequentlyThe action in this semidirect product is the natural evaluation of group automorphisms on $G$. This normalizer is the holomorph, realized as all permutations $x\mapsto t\phi (x)$. The decomposition is established from the regular group action itself rather than only from an order comparison.

Choose $\alpha \in \Omega$ and identify $\Omega$ with $N$ by $n\mapsto n\alpha$. Normality means that the point stabilizer $H_{\alpha }$ acts on $N$ by conjugation, andDouble transitivity therefore says that $H_{\alpha }$ is transitive on $N\setminus \{1\}$. All nonidentity elements of $N$ have the same order. By Cauchy's theorem for finite groups, some element has order a prime $p$ dividing $|N|$, so that common order is $p$. Cauchy's theorem excludes every other prime from $|N|$, hence $N$ is a finite p-group.

By the center argument from Q1, $Z(N)\ne 1$. The center is characteristic in $N$ and therefore invariant under the conjugation action of $H_{\alpha }$. Its nonidentity elements form a nonempty invariant subset of the transitive set $N\setminus \{1\}$, so $Z(N)=N$. Thus $N$ is abelian of exponent $p$:This is elementary abelian regular kernels in doubly transitive groups; knowing only that all element orders agree would not by itself prove commutativity.

Every $h\in H$ is uniquely $n{h}_0$ with $n\in N$ and $h_0\in H_{\alpha }$, because regularity supplies the unique $n$ sending $\alpha$ to $h\alpha$. Hence $H=N⋊H_{\alpha }$. The conjugation action of $H_{\alpha }$ on $N$ is faithful: an element centralizing $N$ and fixing $\alpha$ fixes every $n\alpha$, and is the identity permutation. Group automorphisms of an elementary abelian p-group are exactly invertible ${\mathbb{F}}_p$-linear maps, so $H_{\alpha }\le GL(d,p)$. Choosing a vector-space basis therefore givesUnder this embedding, $N$ is the translation subgroup and the point stabilizer is a linear subgroup transitive on nonzero vectors. The action is assumed to have at least two points, as usual for a doubly transitive group.

A group automorphism is a bijective group homomorphism from $G$ to itself. The multiplication in $\mathrm{Aut}(G)$ is composition: $(\alpha \beta )(g)=\alpha (\beta (g))$. Composition is associative, the identity map is the identity element, and each inverse bijection is again a homomorphism, so these maps form the automorphism group.

For conjugation by $h$, multiplication is preserved becauseConjugation by $h^{-1}$ is its inverse, so $\psi (h)$ is an group automorphism. MoreoverHence $\psi :G\to \mathrm{Aut}(G)$ is a homomorphism, with image the inner automorphisms. Its group homomorphism kernel is the group center of $G$, since $\psi (h)$ is the identity precisely when $hg=gh$ for every $g$.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.

Undo conjugation by $\pi$. The remaining group automorphism $\beta$ fixes each cyclic $3$-subgroup, so it fixes or inverts each generator. For adjacent supports, inverting exactly one generator changes the order of their product between two and three; inverting both preserves that order. Hence the two inversion choices must agree. The triple-support graph is connected, since one can replace differing points one at a time, so the choice is uniform. Inverting every $3$-cycle is impossible: take $c=(123)$, $d=(142)$, whose product is the $3$-cycle $(143)$. Then $\beta (cd)=(cd{)}^{-1}=d^{-1}{c}^{-1}$, whereas the group homomorphism property would give $c^{-1}{d}^{-1}$; these are unequal because $c,d$ do not commute. Therefore $\beta$ fixes every $3$-cycle and is the identity. Conversely, conjugation by $S_n$ acts faithfully on $A_n$, since centralizing all $3$-cycles fixes every triple and every point. The automorphisms of alternating groups from triple supports are exactly

Relabel each duad of each synthematic total by the chosen transposition and compare the resulting synthematic total with the six labelled synthematic totals. For example, $(12)$ interchanges $T_1,T_2$, interchanges $T_3,T_4$, and interchanges $T_5,T_6$. Applying the same calculation to the other adjacent transpositions givesThese formulae use the PDF's numbering of the synthematic totals, which is essential for the particular fixed elements below. Compose permutations from left to right, consistently with the right action on synthematic totals. Multiplying the second and fourth images gives $(23)(45)$, while multiplying the third and fifth gives $(34)(56)$. Hence the two involutionsare fixed by the outer automorphism $\alpha$. Since $\alpha$ is a group automorphism, their product is fixed as well. Applying $u$ and then $v$ yieldsUnder right-to-left composition the product is the inverse cycle, which is also fixed; the displayed convention removes that ambiguity.