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Centralizers in the Symmetric Group

For two different elements $\sigma, \tau\in S_n$ which are non-conjugate it can happen that their centralizers are same (or conjugate subgroups). I learned of this just recently. I worked out and found examples: For the two partitions $n = (n-2) +2$ and $n= (n-2) + 1 + 1$ taking taking elements of the respective conjugacy classes, $\sigma = (1,2,\ldots, n-2)(n-1, n)$ and $\tau =(1,2,\ldots,n-2)$, the centralizers are both same, namely the subgroup generated by $\sigma$.

Is there are any other pair with the same behaviour in $S_n$? For a general group $G$, it is easily seen that if $z\in Z(G)$, then $g$ and $zg$ have the same centralizers. What about groups with trivial centre?

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First, note that, in your example, the centraliser is actually the group generated by $\tau$ and $(n-1,n)$, which, when $n$ is even, is NOT equal to the group generated by $\sigma$. (It's not even cyclic.)

Here is a generalisation of your example: if $A$ is a partition of $n-2$ without any part of size $1$ or $2$, then $[A,2]$ and $[A,1,1]$ will provide an example.

I think you could probably prove that these are the only examples. Here is (maybe) a sketch of a proof. If $\tau$ is a permutation and $C$ is its centraliser, then the orbits of $C$ partition $\{1,\ldots,n\}$ according to the length of the cycle they are contained in $\tau$. Moreover, $C$ is the direct product of its action on each of its orbits. Therefore you can reduce to the case where each permutations has only one length of cycle, and I think it's not too hard to show that the only case that works is $[1,1]$ vs $[2]$.

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