How many distinct ways are there to color the 12 edges of a cube? I know that I will ultimately be using the orbit counting theorem involving $$\frac{1}{|G|}\sum_{g\in G}|\mbox{Fix}_A(g)|$$.  Where specifically for the cube, the group is $S_4$, $A=\{\mbox{colorings of edges of cube with two colors}\}$ and each $g$ (or $w$ as expressed below) is a cycle type conjugacy class representative of the axis of symmetries of a cube (for example, $(2,2)$ is the cycle type representative of a $90^{\circ}$ rotation about the axis connecting two opposite centers of faces).  I know the following about the size of these conjugacy classes:
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{conjugacy class $\mathcal{C}$} & (1,1,1,1) & (2,1,1) & (2,2) & (3,1) & (4)\\ \hline
|\mathcal{C}| & 1 & \binom{4}{2}=6 & \binom{3}{1}=3 & \binom{4}{1} 2!=8 & 3!=6\\ \hline
\end{array} $$
I also know that $|A|=2^{12}$.
What I am primarily having trouble with is understanding which conjugacy class represents which rotation.  Further, then struggling to visualize what those rotations are doing for the edges I am trying to fix.  I am trying to fill out the following table in order to calculate the number of distinct ways:
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{conjugacy class $\mathcal{C}$} & (1,1,1,1) & (2,1,1) & (2,2) & (3,1) & (4)\\ \hline
|\mathrm{Fix}_A(w)| &2^{12}  &  & 2^2 &  & \\ \hline
\end{array} $$
 A: To answer  this question  we first  need the cycle  index of  the edge
permutation group of the cube $E$  by rotations, which we now compute.
There is  the identity,  which contributes $$a_1^{12}.$$  Rotations by
$120$ degrees and $240$ degrees about an axis passing through pairs of
opposite vertices contribute  $$4\times 2 a_3 ^4.$$  Rotations by $90$
and $270$ and  $180$ degrees about an axis passing  through the center
of opposite faces  contribute $$3 \times (2 a_4^3  + a_2^6).$$ Finally
rotations by $180$ degrees about  an axis passing through midpoints of
opposite edges contribute $$6\times a_1^2 a_2^5.$$

This gives the cycle index
$$Z(E) = \frac{1}{24}
\left(a_1^{12} + 8 a_3^4 + 6 a_4^3 + 3 a_2^6 + 6 a_1^2 a_2^5\right).$$
With this  cycle index we obtain  by Burnside for edge  colorings with
$n$ colors the formula
$$\frac{1}{24} \left(n^{12} + 8 n^4 + 6  n^3 + 3 n^6 + 6 n^7\right).$$
This gives the  sequence $$1, 218, 22815,  703760, 10194250, 90775566,
576941778,    2863870080,\ldots$$   which    points   us    to   OEIS
A060530, where we find  that indeed we have
the right cycle index. For two colors we obtain the value
$$\bbox[5px,border:2px solid #00A000]{
218.}$$
