Conjugacy class of transpositions is preserved by automorphisms

For the symmetric group on a finite set
Suppose $$S$$ is a finite set of size greater than $$1$$ and not equal to $$6$$ and $$G = \operatorname{Sym}(S)$$ is the symmetric group on $$S$$. Then, the conjugacy class of transpositions in $$G$$ is preserved by automorphisms. In other words, any automorphism of $$G$$ sends transpositions to transpositions.

For the symmetric group on an infinite set
Suppose $$S$$ is an infinite set and $$G = \operatorname{Sym}(S)$$ is the symmetric group on $$S$$. Then, the conjugacy class of transpositions in $$G$$ is preserved by automorphisms of $$G$$. In other words, every automorphism of $$G$$ sends transpositions to transpositions.

Facts used

 * 1) uses::Conjugacy class of transpositions is the unique smallest conjugacy class of involutions: This statement holds when the size of the set is not equal to $$6$$.
 * 2) uses::Finitary alternating group is monolith in symmetric group

A proof that works only for the finite case
For the finite case, fact (1) gives the proof. This is because any automorphism preserves the order of elements, and preserves conjugacy, hence it permutes the conjugacy classes of involutions (i.e., of elements of order two). Since the conjugacy class of transpositions is the unique smallest conjugacy class of involutions, it cannot be mapped to another conjugacy class of involutions, hence it remains invariant under automorphisms.

A proof that works in general
Given: A set $$S$$ of size greater than $$1$$ and not equal to $$6$$ (possibly infinite). $$G$$ is the symmetric group on $$S$$. A transposition $$\sigma$$ of $$S$$ and an involution $$\tau$$ of $$S$$ that is not a transposition.

To prove: There is no automorphism of $$S$$ sending $$\tau$$ to $$\sigma$$.

Proof: If there were an automorphism of $$S$$ sending $$\tau$$ to $$\sigma$$ it would send $$C_G(\tau)$$ to $$C_G(\sigma)$$, so the two groups would be isomorphic, and hence, the largest normal $$2$$-subgroup of $$C_G(\sigma)$$ is isomorphic to the largest normal $$2$$-subgroup of $$C_G(\tau)$$. (By a $$2$$-subgroup, we mean a subgroup where all the elements have order a power of $$2$$).

Suppose $$\sigma = (i,j)$$ and $$T$$ is the complement of $$\{ i,j \}$$. Then, $$C_G(\sigma) = \langle \sigma \rangle \times \operatorname{Sym}(T)$$.

We first consider the case that $$S$$ has at least seven elements, so that $$T$$ has at least five elements.

For finite $$T$$ fact (2) yields that if $$T$$ has five or more elements or is infinite, then the finitary alternating group on $$T$$ is contained in every nontrivial normal subgroup of $$\operatorname{Sym}(T)$$. From this, we see that $$\operatorname{Sym}(T)$$ has no nontrivial normal $$2$$-subgroup. Thus, the largest normal $$2$$-subgroup of $$C_G(\sigma)$$ is $$\langle \sigma \rangle$$.

On the other hand, $$\tau$$ is a product of (possibly infinitely many) disjoint transpositions. The subgroup generated by all these transpositions is a normal $$2$$-subgroup in $$C_G(\tau)$$. Since $$\tau$$ is not a transposition, it involves at least two transpositions, so $$C_G(\tau)$$ has a normal $$2$$-subgroup of order at least four. Thus, the largest normal $$2$$-subgroup of $$C_G(\sigma)$$ is not isomorphic to that of $$C_G(\tau)$$.

We now consider the case that $$S$$ has fewer than six elements. If $$S$$ has two elements, three elements, or five elements, the largest normal $$2$$-subgroup of $$T$$ is still trivial, so the above applies.

When $$S$$ has four elements, $$T$$ has two elements, and in this case, $$C_G(\sigma)$$ is a Klein four-group. The only other possible conjugacy class of involutions is double transpositions, and for a double transposition $$\tau$$, $$C_G(\tau)$$ is isomorphic to the dihedral group of order eight.