Automorphism group of finitary symmetric group equals symmetric group

Statement
Let $$S$$ be an infinite set. Let $$K = \operatorname{Sym}(S)$$ be the fact about::symmetric group on $$S$$, i.e., the group of all permutations on $$S$$, and $$G = \operatorname{FSym}(S)$$ be the subgroup of $$K$$ that comprises the finitary permutations. In other words, $$G$$ is the fact about::finitary symmetric group on $$S$$.

Then, $$G$$ is normal in $$K$$. Further consider the map $$\varphi: K \to \operatorname{Aut}(G)$$ sending an element $$a \in K$$ to conjugation by $$a$$ on $$G$$. This map is an isomorphism.

Related facts

 * Symmetric groups on infinite sets are complete
 * Automorphism group of finitary alternating group equals symmetric group

Facts used

 * 1) uses::Finitary symmetric group is normal in symmetric group
 * 2) uses::Finitary symmetric group is centralizer-free in symmetric group
 * 3) uses::Conjugacy class of transpositions is preserved by automorphisms
 * 4) uses::Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation

Proof

 * 1) By fact (1), the map $$\varphi: K \to \operatorname{Aut}(G)$$ is well-defined, and by fact (2), $$\varphi$$ is injective.
 * 2) By fact (3) every element of $$\operatorname{Aut}(G)$$ sends transpositions to transpositions. Combining this with fact (4), we obtain that every element of $$\operatorname{Aut}(G)$$ comes from an inner automorphism in $$K$$. Thus, $$\varphi$$ is surjective.