Let $\sigma=(1,2,11)(3,4)(5,6,7,8,9)\in S_{12}$. I am willing to get the number of conjugates of $\sigma$. Clearly if $\tau$ be one such, then it must have the same cycle type. So in other words, we need to search for number of elements of cycle type $[2,3,4]$ in $S_{12}$.
And the number is $\binom{12}{2}\binom{10}{3}\binom{7}{5}$.
Please tell me if I made any mistake. Thank you in advance
The number $$\binom{12}{2}\binom{10}{3}\binom{7}{5}$$ counts the ways to pick numbers that belong to each cycle. We must also specify how the numbers are arranged in the cycle. The number of ways to arrange $k$ numbers is $k!$, but cyclic permutation of an arrangement does not change the $k$-cycle, so since there are $k$ cyclic permutations of $k$ numbers we have (by the orbit-stablizer theorem or otherwise) that the number of distinct arrangements of the elements in a $k$-cycle is $$\frac{k!}{k}=(k-1)!$$ We must therefore multiply by $(2-1)!(3-1)!(5-1)!$ to obtain $$\binom{12}{2}\binom{10}{3}\binom{7}{5}2!4!=7983360$$