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The set $S(\mathbb{N})$ consists of all bijections $\mathbb{N}\to \mathbb{N}$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb{N})$ is a group.

An element of $S_{\mathrm{fin}}(\mathbb{N})$ has finite support. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb{N})$ is a subgroup.

Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequencemust repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so ${\sigma }^l$ fixes every point. No smaller positive power fixes $n$. Thereforewhich is finite.

For $\tau \in S_{\mathrm{fin}}(\mathbb{N})$, partition its finite support into the orbits of the cyclic group $⟨\tau ⟩$. On each orbit, let ${\sigma }_i$ agree with $\tau$ and fix every point outside that orbit. Then each ${\sigma }_i$ is a permutation cycle, their supports are pairwise disjoint, andDisjoint cycles commute because at every point at most one of them acts nontrivially.

Writing $l_i=\mathrm{ord}({\sigma }_i)$, a power ${\tau }^m$ is the identity exactly when every ${\sigma }_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus

For $n\ge 2$ with $n\ne 6$, bijectivity of $\varphi$ and the previous part implyThe supplied numerical fact forces $k=1$. Thus $\varphi$ permutes the transpositions.

Suppose $n\ge 3$, and write $s_r=\varphi ((1r))$ for $r=2,\dots ,n$. These are distinct transpositions. For $r\ne t$, the product $(1r)(1t)$ has order of a group element three, so $s_r{s}_t$ also has order three. Two distinct transpositions have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element support of a permutation associated with each $s_r$ therefore meets that of every other $s_t$.

We now show that they have a common point. Choose two; the support of a permutation is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.

For $n\ge 5$, there are $n-1\ge 4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper subgroup of $S_4$, contrary to surjectivity of $\varphi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge 3$ there is a common point $a$, and we may writeThe $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a permutation $\sigma$ byThenThe two group homomorphisms $\tau \mapsto \varphi (\tau )$ and $\tau \mapsto {\sigma }^{-1}\tau \sigma$ agree on a generating set of a group, so agree on every product of its elements, hence throughout $S_n$:Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every automorphism of either is the identity and we can take $\sigma =e$. This completes all degrees $n\ne 6$ and proves that every such automorphism is an inner automorphism.

A positive bump of the real line is an order automorphism whose nonempty support of a permutation is one open interval and which moves every point of that interval to the right. It fixes the complement. Iterating any interior point tends to the right endpoint in positive time and to the left endpoint in negative time.