Finitary symmetric group is not injective endomorphism-invariant in symmetric group

Statement
Suppose $$S$$ is an infinite set. Denote by $$G = \operatorname{Sym}(S)$$ the symmetric group on $$S$$ and by $$H = \operatorname{FSym}(S)$$ the finitary symmetric group on $$S$$, viewed as a subgroup of $$G$$.

Then, $$H$$ is not an disproves property satisfaction of::injective endomorphism-invariant subgroup of $$G$$. In other words, there exists an injective endomorphism $$\sigma$$ of $$G$$ such that $$\sigma(H)$$ is not contained in $$H$$.

Related facts

 * Finitary symmetric group is characteristic in symmetric group

Axiomatic assumptions
For certain cardinalities of $$S$$ (including the countably infinite cardinality), the proof does not rely on the axiom of choice. However, the assertion for all infinite $$S$$ does rely on the axiom of choice.

Proof
We construct a bijection $$\alpha$$ between $$S \times \mathbb{N}$$ and $$S$$. The bijection is explicit for some infinite cardinalities of $$S$$, and its existence is guaranteed by infinite cardinal arithmetic for all infinite $$S$$ by the axiom of choice.

We now consider the composite mapping:

$$\operatorname{Sym}(S) \to \operatorname{Sym}(S \times \mathbb{N}) \to \operatorname{Sym}(S)$$

The first homomorphism simply operates by acting only on the first of two coordinates, ignoring the second coordinate. The second homomorphism operates using the bijection $$\alpha$$. Let $$\sigma$$ be the composite mapping. The following are easy to check:


 * $$\sigma$$ is injective: The first map in the composite is injective, and the second map is bijective, so the composite is injective.
 * $$\sigma$$ does not preserve $$H$$: Any element of $$H$$ with nontrivial support gets sent to something that moves infinitely many elements.