= Solution
An <group automorphism> preserves element orders and sizes of <conjugacy classes>. A nonidentity involution in $S_n$ has <cycle type> $2^k1^{n-2k}$ and class size
$$
\frac{n!}{2^kk!(n-2k)!},\qquad 1\leq k\leq\lfloor n/2\rfloor.
$$
For its class to have the same size as the <transposition> class, it must satisfy
$$
\frac{(n-2)!}{(n-2k)!}=2^{k-1}k!.
$$
For fixed $k\geq2$ 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 $4!=2^2\cdot3!$, and thereafter the left side is larger. For $k\geq4$ 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 $n\ne6$, 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 $n\geq5$, the stars have size $n-1\geq4$ 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 $(i\ j)$, 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 $(i\ j)=(a\ i)(a\ j)(a\ i)$ 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). Thus
$$
\boxed{\operatorname{Out}(S_n)=1\quad\text{for }n\ne6.}
$$
The small-degree adjustments are necessary: a cardinality-only star argument would not cover $n=4$.
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