Question from Artin's Algebra:
Let $q$ be a 3-cycle in $S_n$. How many even permutations $p$ are there such that $pqp^{-1}=q$?
Since $q$ is a 3-cycle, I know that it can be written as a product of two transpositions. Thus, $q$ is an even permutation. (I'm not sure if this is helpful or not but it is a fact that I noticed.) The question essentially asks how many even permutations commute with $q$. I have seen that the answer should be $3|A_{n-3}|$ from another question on this site, but I don't understand why.
There are $\frac {n!}{2}$ even permutations in $S_n$
$\frac {(n-3)!}{2}$ will be completely independent from $q$
and if they are independent they commute and $p^{-1} q p = q$
and $(qp)^{-1}q(qp) = p^{-1} q^{-1}qqp = p^{-1} q p$ and $(q^{-1}p)^{-1}q(q^{-1}p) = p^-1 qqq^{-1}p = p^{-1} q p$
The following different idea might be easier:
Calculate the result for $n<5$ by hand (as in these cases there are several classes of $3$-cycles).
If $n>5$, then all $3$-cycles are conjugate under $A_n$, and there are $2{n\choose 3}$ such $3$ cycles. The order of the conjugation stabilizer (=centralizer) of one thus is by the orbit-stabilizer theorem $|A_n|=n!/2$, divided by the orbit (= conjugacy class) length.
Thus the centralizer order is $$ \frac{n!}{2\cdot 2\cdot {n\choose 3}} =\frac{n!\cdot (n-3)! 3!}{4 n!} =\frac{3\cdot (n-3)!}{2} $$
as attributed to another posting.