Here is a question that I am stuck at:
What is the minimum size of the centralizers of elements in $S_n$?
It is known to me that for a permutation $\sigma\in S_n$ if $(a_1,a_2,\dots,a_n)$ denote the frequency of $i-$cycles of $\sigma$ (meaning there are $a_i$ many $i-$cycles), then $|C(\sigma)|=\prod_{j=1}^{n}(j^{a_j})(a_j!)=(1^{a_1}2^{a_2}\cdots n^{a_n})(a_1!a_2!\cdots a_n!)$ (this is also an exercise in Dummit and Foote).
So the idea must be to minimize $\prod_{j=1}^{n}(j^{a_j})(a_j!)$ subject to the condition $a_1+2a_2+\cdots+na_n=n$. Again here I have seen some obscure comments in MO that suggested that the maximum size of a conjugacy class, that of $(1\,2\,\cdots\, n-1)(n)$, is $n!/(n-1)$. This is what is obtained by taking $a_1=1=a_{n-1}$. The way I guessed it was purely by hit or miss.
I thought ideally I would like if all $a_i$'s are 0 but this is disallowed by the sum condition. So next we try to have only $1$'s as the factorial terms and based on that control the powers $j^{a_j}$'s and with the cycle type $(n-1,1)$ I came up with the minimum value.
I guess my intuition is correct, but $\textit{I am unable to write a proof nicely}$.
Also a second question which I guess would be naturally more difficult to answer than the first one :
Is there a definitive way of finding two distinct sequences for distinct cycle types like $(a_1,...,a_n)$ and $(b_1,...,b_n)$ with $\sum_{j=1}^{n}ja_j=\sum_{j=1}^{n}jb_j=n$ such that $\prod_{j=1}^{n}(j^{a_j})(a_j!)=\prod_{j=1}^{n}(j^{b_j})(b_j!)$?
The reason to ask that question is that other that $S_3$ we can always find two or more same summands in the class equation of $S_n$ for $n\geq 4$ (example $1+3+6+6+8$ for $S_4$).
Thanks in advanced for responses.
The first question is elementary number theory. Let $g \in S_n$, let $k_1$ be the number of fixed points of $g$ (cycles of length 1), and let the lengths of the remaining cycles of $g$ be $k_2,k_3,\ldots,k_r$.
Then $k_1 + \cdots +k_r = n$, and the centralizer of $g$ in $S_n$ has order at least $k_1k_2 \cdots k_r$, which is at least $n-1$ in all cases.
Furthermore, $k_1=1$, $k_2=n-1$ is the only way to get $n-1$ except for $1+2+2$ with $n=5$. But in fact the centralizer of $(1,2)(3,4)$ in $S_5$ has order $8$.
The second question is more open ended. The equal centralizers for small values of $n$ immediately give rise to examples for larger $n$. For example, the equality for $(1,1,2)$ and $(4)$ in $S_4$, gives rise to example $(1,1,2,n-4)$ and $(4,n-4)$ (and many more) for larger $n$. So I guess you could ask are all examples in some sense induced from examples with small $n$ or are there genuinely different types of examples for larger $n$. I don't know!