= Solution
{c}
<Conjugation> of a <permutation> relabels each symbol by the inverse conjugating permutation. Specifically,
$$
\boxed{\tau^{-1}(a_1a_2\cdots a_n)\tau
=(\tau^{-1}(a_1)\ \tau^{-1}(a_2)\ \cdots\ \tau^{-1}(a_n)).}
$$
To verify the direction of the inverse, apply the left side to $\tau^{-1}(a_j)$: $\tau$ first takes it to $a_j$, the original cycle takes it to $a_{j+1}$, and $\tau^{-1}$ takes it to $\tau^{-1}(a_{j+1})$, with cyclic indexing. Any symbols fixed by the original <permutation> are similarly relabelled and remain fixed. More generally, conjugate every cycle of a disjoint-cycle decomposition in this way; disjointness is preserved because $\tau^{-1}$ is a <bijection>.
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