**Context**: It is well known that given a permutation in $S_n$ with $a_i$ $i$-cycles (when written as a product of disjoint cycles), the size of the conjugacy class is given by 
$$ \frac{n!}{\prod_{j=1}^n (j)^{a_j}(a_j !)}$$

It is also known that the conjugacy class containing permutations with exactly one fixed point and exactly one $n-1$ cycle (hence, $a_1 = 1$ and $a_{n-1} = 1$) has maximum conjugacy class size.
(see this [post][1])


**Question:** Now let's fix the number of cycles, $m$ (this count includes the trivial ones). Then how would one go about finding the cycle type with the maximum conjugacy class size, among the ones with $m$ cycles?

Note that the case for $m=2$ is true for $n \geq 3$ by the assertion above in the context.

My "conjecture" is that any such cycle type needs to have $a_1 \geq \frac{m-1}{2}$. (i.e. at least about half of the cycles are trivial). And that about a fourth are 2 cycles and so on.






  [1]: https://mathoverflow.net/questions/2888/injective-proof-about-sizes-of-conjugacy-classes-in-s-n