Talk about conjugacy classes in the symmetric group \(S_n\).
Solution
1. Two permutations in \(S_n\) are conjugate if and only if they have the same cycle type.
If \[ \sigma=(a_{11}\ \cdots\ a_{1r_1})\cdots(a_{k1}\ \cdots\ a_{kr_k}), \] then for any \(g\in S_n\), \[ g\sigma g^{-1}=(g(a_{11})\ \cdots\ g(a_{1r_1}))\cdots(g(a_{k1})\ \cdots\ g(a_{kr_k})), \] so conjugation preserves cycle lengths.
Conversely, if two permutations have the same multiset of cycle lengths, choose a bijection sending the entries of each cycle of the first permutation, in cyclic order, to the entries of a cycle of the same length in the second. Extending these bijections over all cycles gives \(g\in S_n\) conjugating one permutation to the other.
2. Hence conjugacy classes in \(S_n\) are indexed by partitions of \(n\).
The cycle lengths of a permutation form a partition \[ n=1^{m_1}2^{m_2}\cdots n^{m_n}, \qquad \sum_i i m_i=n. \] By 1, this partition determines the conjugacy class and every partition occurs as a cycle type.
3. If a permutation has cycle type \(1^{m_1}2^{m_2}\cdots n^{m_n}\), then its conjugacy class has size\[ \frac{n!}{\prod_{i=1}^n i^{m_i}m_i!}. \]
Its centralizer has order \[ \prod_{i=1}^n i^{m_i}m_i!. \] Indeed, for each length \(i\), one may independently rotate each of the \(m_i\) cycles in \(i^{m_i}\) ways and permute those equal-length cycles in \(m_i!\) ways. Orbit-stabilizer for the conjugation action then gives the stated class size.