Order of the centralizer of a permutation

Let $n_1,n_2,\ldots,n_k$ be the distinct lengths of the cycles of $\sigma$ (including 1 if there are fixed points) and suppose that there are $m_i$ cycles of length $n_i$. Then the centralizer of $\sigma$ can permute the cycles of the same length. Its order is $\prod_{i=1}^k n_i^{m_i}m_i!$.

Calculating the centralizer of a subgroup $H$ of $S_n$ is not difficult, but it is more complicated. The order of the centralizer of a single orbit is equal to the number of fixed points (in that orbit) of the stabilizer of a point in the orbit. But if $H$ has more than one orbit with equivalent actions then the equivalent orbits can be permuted by the centralizer, so the complete centralizer is a direct product of wreath products of centralizers of sets of equivalent orbits.

You let the permutations act by conjugation on the permutation and you seek the size of the stabilizer of $\sigma$. By the orbit-stabilizer theorem, it is enough to know the size of the orbit, which is the well-known size of the conjugacy class of $\sigma$.