Conjugate to the Permutation How many elements in $S_{12}$ are conjugate to the permutation $$\sigma=(6,2,4,8)(3,5,1)(10,11,7)(9,12)?$$
How many elements commute with $\sigma$ in $S_{12}$?
I believe I use the equation $n!/|K_{\sigma}|$ for the second question, but I'm not sure. Is anyone aware of how to do these?

Two permutations are conjugate if they have the same cycle structure (same number of cycles with same lengths).
Given a permutation with this cycle structure, you won't change it if you rotate each cycle as much as you want.
So, there are $4$ "rotated cycles" for the first one, $3$ for the next two cycles, and $2$ for the last. So there are $4\times3\times3\times2$ equivalent permutations. But you can also swap the second and the third cycles, so there are actually $(4\times3\times3\times2)\times2=144$ equivalent permutations, for each given permutation.
So there are $12!/144= 3326400$ permutations conjugate to yours.

A precision about what I mean by "rotated cycle".
Since you have
$$\tau^{-1}\sigma\tau = \left(\tau(6), \tau(2), \tau(4), \tau(8)\right)\left(\tau(3), \tau(5), \tau(1)\right)\left(\tau(10), \tau(11), \tau(7)\right)\left(\tau(9), \tau(12)\right)$$
A priori all permutation $\tau \in S_n$ would give you a conjugate of $\sigma$ by this operation, but there are double counts.
If, for the first cycle, the image of $(\tau(6), \tau(2), \tau(4), \tau(8))$ is, say, $(1,2,3,4)$, then you will get the same cycle if instead the image is $(2,3,4,1)$, $(3,4,1,2)$ or $(4,1,2,3)$. So each given conjugate can be associated with $4$ permutations $\tau$ that give the same cycle $(1,2,3,4)$. But you have to consider also the other cycles. And you need also to consider that the two 3-cycles may be switched.
