This is a way to approach this problem.

Let $\displaystyle G$ be the group of all trasformations of the octahedron. In the other thread it was shown that $\displaystyle |G| = 48$. And $\displaystyle X$ set of all colorings (even the possibly similar one).

Now put these symettries into conjugacy classes which can be found

here. For each conjugacy class, say 90 degree rotation, find which elements in $\displaystyle X$ are fixed and how many of them. This number gets multipled by six since there are 90 degree rotations. And go through each class. After you done the hard part the rest follows by

Burnside's lemma - be sure to look at the example with the cube in that link, it is very similar.