Symmetric group on infinite coinfinite subset is not conjugacy-closed

Statement
Suppose $$S \subset T$$ are sets such that both $$S$$ and $$T \setminus S$$ are infinite. Let $$\operatorname{Sym}(S), \operatorname{Sym}(T)$$ denote the symmetric groups on $$S$$ and $$T$$ respectively, with $$\operatorname{Sym}(S)$$ embedding naturally in $$\operatorname{Sym}(T)$$: any permutation on $$S$$ is extended to a permutation on $$T$$ by fixing all elements of $$T \setminus S$$.

With this embedding, $$\operatorname{Sym}(S)$$ is not a conjugacy-closed subgroup of $$\operatorname{Sym}(T)$$.

Proof
(This proof assumes the axiom of choice).

Given: $$S \subset T$$ such that both $$S$$ and $$T \setminus S$$ are finite.

To prove: The symmetric group $$\operatorname{Sym}(S)$$ is not conjugacy-closed in $$\operatorname{Sym}(T)$$. In other words, there exist elements $$g,h \in \operatorname{Sym}(S)$$ such that $$g$$ and $$h$$ are not conjugate in $$\operatorname{Sym}(S)$$ but are conjugate in $$\operatorname{Sym}(T)$$.

Proof:


 * 1) Let $$A$$ be a countably infinite subset of $$S$$ such that $$S \setminus A$$ have the same size (this can be done if $$S$$ is infinite). Also note that there is a bijection between $$T \setminus S$$ and $$A \cup (T \setminus S)$$. Combining these two, we get a bijection $$\alpha: S \to S$$ such that $$\alpha(S) = S \setminus A$$ and $$\alpha(T \setminus S) = A \cup (T \setminus S)$$.
 * 2) Suppose $$f$$ is a permutation of $$S$$ that moves every element of $$S$$. Such a permutation exists: for instance, we can use the axiom of choice to partition $$S$$ into subsets of size two, and then select a permutation that interchanges the two elements in any subset.
 * 3) Consider the permutations $$f$$ and $$\alpha f \alpha^{-1}$$. $$f$$ is a permutation that moves every element of $$S$$ and fixes all elements of $$T \setminus S$$, while $$\alpha f \alpha^{-1}$$ moves every element of $$S \setminus A$$ and fixes every element of $$A \cup (T \setminus S)$$. We note that:
 * 4) * $$f$$ and $$\alpha f \alpha^{-1}$$ are not conjugate in $$\operatorname{Sym}(S)$$, because $$\alpha f \alpha^{-1}$$ has fixed points but $$f$$ doesn't have fixed points.
 * 5) * $$f$$ and $$\alpha f \alpha^{-1}$$ are conjugate in $$\operatorname{Sym}(T)$$ by $$\alpha$$.