Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation

Statement
Let $$S$$ be a set, $$K = \operatorname{Sym}(S)$$ be the fact about::symmetric group on $$S$$, and $$G = \operatorname{FSym}(S)$$ be the subgroup of $$G$$ comprising the finitary permutations, i.e., $$H$$ is the fact about::finitary symmetric group on $$S$$. Suppose $$\sigma$$ is an automorphism of $$H$$ that sends transpositions to transpositions. Then, $$\sigma$$ is induced by conjugation by some element of $$G$$.

Related facts

 * Transposition-preserving automorphism of symmetric group is inner: This is the version for finite sets.

Stronger facts

 * Symmetric groups on finite sets are complete: This holds except when the set has size two or six.
 * Symmetric groups on infinite sets are complete

Facts used

 * 1) uses::Transpositions generate the finitary symmetric group
 * 2) uses::Transpositions commute iff they are disjoint

Proof
Given: A finite set $$S$$. $$G$$ is the finitary symmetric group on $$S$$. $$\sigma$$ is an automorphism of $$G$$ that sends transpositions to transpositions.

To prove: $$\sigma$$ is induced by conjugation by a permutation in $$K$$, the whole symmetric group on $$S$$.

Proof: By fact (1), it suffices to find $$g \in G$$ such that conjugation by $$g$$ agrees with $$\sigma$$ on the set of transpositions. Further, we can assume that $$S$$ has at least three elements (The statement is obviously true for $$S$$ having zero, one, or two elements). We now describe how the permutation $$g$$ can be constructed explicitly.

Also, we use fact (2) at many steps, without explicitly acknowledging it.


 * 1) Pick distinct elements $$a,b,c \in S$$. Then, both $$\sigma((a,b))$$ and $$\sigma((a,c))$$ are transpositions. We claim that these transpositions have exactly one element in common: If $$\sigma((a,b))$$ and $$\sigma((a,c))$$ have no element in common, then they commute, and hence, $$(a,b)$$ commutes with $$(a,c)$$, which is not true. Thus, $$\sigma((a,b))$$ and $$\sigma((a,c))$$ have exactly one element in commmon.
 * 2) Pick distinct elements $$a,b,c \in S$$. Then, there are elements $$a',b',c' \in S$$ such that $$\sigma((a,b)) = (a',b')$$, $$\sigma((b,c)) = (b',c')$$ and $$\sigma((c,a)) = (c',a')$$: This follows directly from the previous step.
 * 3) For any element $$a \in S$$, there is a unique element $$g(a)$$ that is involved in the transposition $$\sigma((a,b))$$ for every $$b \ne a$$: For any $$b \ne c$$ distinct from $$a$$, there exist $$a',b',c'$$ as described in the previous step. We now show that $$\sigma((a,d))$$ is also a transposition involving $$a'$$.
 * 4) * Since $$(a,d)$$ commutes with $$(b,c)$$, $$\sigma((a,d))$$ commutes with $$(b',c')$$. In particular, it cannot be a transposition involving $$b'$$ or $$c'$$.
 * 5) * Since $$(a,b)$$ does not commute with $$(a,d)$$, $$\sigma((a,b))$$ does not commute with $$\sigma((a,d))$$, so they must have an element in common. Thus, either $$a'$$ or $$b'$$ is involved in $$\sigma((a,d))$$.
 * 6) * These together force $$a'$$ to be involved in $$\sigma((a,d))$$.
 * 7) * We can thus define $$g(a)$$ as the unique element involved in $$\sigma((a,x))$$ for every transposition involving $$a$$.
 * 8) The previous step yields a map $$g:S \to S$$ such that for any transposition $$(a,b)$$, $$\sigma((a,b))$$ is a transposition involving both $$g(a)$$ and $$g(b)$$. Step (2) makes it clear that $$g(a) \ne g(b)$$, so $$\sigma((a,b)) = (g(a),g(b))$$. Thus, $$\sigma$$ is induced by conjugation by the permutation $$g$$.