 Solution

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.