How do I find he conjugacy class of a specific permutation for another permutation In the symmetric group $S_{13}$ how do I find all the permutations $\tau$ that:
$$\tau\alpha\tau^{-1}=\gamma$$
Where $\alpha = (1,2,4,8)(3,5,7,9,11,13)(6,12)$ and $\gamma = (1,2,3,4)(5,6,7,8,9,10,11)(12,13)$
I thought firstly that all the permutations are standing that condition, because $o(\alpha)=o(\gamma)=12$ and that's why $\tau^{12}\alpha^{12}\tau^{{12}^{-1}}=\gamma^{12}$ and that's true for any $\tau$. But there is a smaller number which is the answer. Correct me what wrong in my last conclusion and how to find the exact number of such permutations ? 
Generally, I just know, how to calculate conjugacy class in the whole symmetric group (the number of k-permutations with the same structure) and the conjugacy class of permutation in sub-group of $S_{13}$. I never knew that there is a method to calculate how much permutations are a conjugacy to specific permutation, if you understand, what I mean (like in this example).
 A: Assumption: $\alpha = (1,2,4,8)(3,5,7,9,11,13)(6,12)$ and $\gamma = (1,2,3,4)(5,6,7,8,9,11)(12,13)$ where we changed the cycle structure of $\gamma$ such that both $\alpha$ and $\gamma$ are odd permutations.  Then,
$$\text{Number of $\tau$ that conjugates $\alpha$ to $\gamma$} = \frac{\text{Order of group $S_{13}$}}{\text{Size of conjugacy class for element $\alpha$}}.$$
(I think such a relation holds.  I need to learn more group theory to know why this indeed holds.)
The order of group $S_{13}$ is $13! = 6,227,020,800$ and the next critical step is to calculate the size of conjugacy class containing element $\alpha$.  This requires some combinatorics.
How to find the number of permutations that contain 3 disjoint cycles of length 4, 6 and 2?
$$\left[\binom{13}{4}(4-1)!\right]\cdot \left[\binom{13-4}{6}(6-1)!\right]\cdot \left[\binom{13-4-6}{2}(2-1)!\right] = 129,729,600.$$
The process is as follows:


*

*Take a combination of 4 distinct vertices out of the 13 available vertices.

*There are $(4-1)!$ ways to write a cycle of length 4 using these 4 vertices.

*Take a combination of 6 distinct vertices out of the remaining 9 vertices.

*There are $(6-1)!$ ways to write a cycle of length 6 using these 6 vertices.

*Take a combination of 2 distinct vertices out of the remaining 3 vertices.

*There are $(2-1)!$ ways to write a cycle of length 2 using these 2 vertices (which is only 1 way of doing so).


Therefore, given $\alpha$ and a certain $\gamma$ in the conjugation class of $\alpha$, the number of permutations $\tau$ that can specifically conjugate $\alpha$ to $\gamma$ is
$$\frac{13!}{\left[\binom{13}{4}3!\right]\cdot \left[\binom{9}{6}5!\right]\cdot \left[\binom{3}{2}1!\right]} = 48.$$
