= Solution
{c}
Conjugating a <cycle> simply relabels its entries. With <permutations> composed from right to left,
$$
\rho(a_1\ a_2\ \cdots\ a_\ell)\rho^{-1}
=(\rho(a_1)\ \rho(a_2)\ \cdots\ \rho(a_\ell)).
$$
This follows by applying each side to $\rho(a_j)$ and to points outside the displayed <cycle>. Consequently <conjugation> preserves the lengths in the <disjoint cycle decomposition>, including fixed points, and therefore preserves <cycle type>.
Conversely, suppose two <permutations> have the same <cycle type>. Pair their <cycles> of equal length, including their one-cycles. Define a <bijection> $\rho$ by mapping the entries of each <cycle> of the first <permutation>, in their cyclic order, to the entries of the paired <cycle> of the second. Because the <cycles> partition $\{1,\ldots,n\}$, these assignments combine into a <permutation>. The displayed identity then conjugates the first <permutation> to the second. Thus they are <conjugate group elements> exactly when their <cycle types> agree.
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