Find all permutations that commute with $\omega$=(1 9 7 10 12 2 5)(4 11)(3 6 8) in $S_{12}$ I'm asked in this exercise to find all permutations that commute with $\omega$=(1 9 7 10 12 2 5)(4 11)(3 6 8) in $S_{12}$. 
What I have so far:
We could write $x$(1 9 7 10 12 2 5)(3 6 8)=(1 9 7 10 12 2 5)(3 6 8)$x$, which means (1 9 7 10 12 2 5)(4 11)(3 6 8)=$x^{-1}$(1 9 7 10 12 2 5)(4 11)(3 6 8)$x$=(1x 9x 7x 10x 12x 2x 5x)(4x 11x)(3x 6x 8x). This means $x$ commutes with $\omega$ iff $x$ 'transfers' one of the distinct cycles that construct $\omega$ unto itself. 
There are $7\cdot 2\cdot 3$ ways to 'present' $\omega$ disregarding the order of multiplication of the cycle (should I disregard it?), each of which creates a distinct commuting permutation if constructed by the algorithm: 1x$\rightarrow$(1 9 7 10 12 2 5), 9x$\rightarrow$(1 9 7 10 12 2 5), et cetera. So all in all we end up with $7\cdot 2\cdot 3$ permutations.
I guess what I should ask is: (a) does this sound correct? (b) since the question wants us to find all permutations and not count them, perhaps there is a more general 'form' for the commuting permutations?
 A: Your reasoning is fine, in my opinion.
$\sigma\in S_{12}$ commutes with $\omega$ if and only if $\sigma\omega\sigma^{-1} = \omega$.
Since there are $7\cdot 2\cdot 3$ ways to fix a representation of $\omega$ in the form
$$
\sigma\omega\sigma^{-1} = (\sigma(1)\  \sigma(9)\  \sigma(7)\  \sigma(10)\  \sigma(12)\ \sigma(2)\  \sigma(5))\ (\sigma(4)\  \sigma(11))\ (\sigma(3) \ \sigma(6)\  \sigma(8)),
$$
and each such representation uniquely determines $\sigma$, there are exactly $7\cdot 2\cdot 3$ elements $\sigma\in S_{12}$ which commute with $\omega$.

Here is an alternative solution using the theory of group actions to determine the number of elements of $S_{12}$ which commute with $\omega$:
The question asks for the centralizer $C(\omega)$ of $\omega$ in $S_{12}$.
Now look at the group operation of $G$ on itself by conjugation. Then $C(\omega)$ is the stabilizer of $\omega$. The orbit $O$ of $\omega$ is the conjugacy class of $\omega$ in $S_{12}$, which is the set of all elements of the same cycle type.
The standard counting method for permutations yields
$$\left|O\right| = \frac{12!}{2\cdot 3\cdot 7}.$$
Now by the orbit-stabilizer-theorem, $\left|C(\omega)\right| = \left|S_{12}\right|/\left|O\right| = 2\cdot 3\cdot 7$.
