I am attempting to figure out how many ways there are to draw the figure Z (or any rotation or reflection of it) on each of the six faces of a cube. That is, there are four possible labels for each face: Z, N, И, and mirrored Z.
When counting the number of labelings, any two cube-labelings that are related by a proper rotation of the cube should be considered identical.
The figure Z shares 180-degree rotational symmetry with the cube, and no other symmetry.
I am familiar with Burnside's counting lemma, and I have determined, for example, that if we ignored rotations of Z and instead just counted whether we were using [a rotation of] Z or its mirror image, we would have 10 possible combinations.
But I do not know how to extend this counting procedure in the case where the labels we're assigning to faces also have their own symmetry properties. I wrote a computer program to count them and got 216, but there is always a possibility I made a mistake --- I would much rather know formally how to prove it.
Since there are $6$ faces on the cube and $4$ distinct labelings of each face, there are initially $4^6 = 4096$ labelings. We now have to let the group of symmetries of the cube act on it by proper rotations and enumerate the orbits.
As you mentioned, according Burnside (really Frobenius or Cauchy), we have to consider the fixed-set for each of the $|S_4| = 4! = 24$ elements of the permutation group of the cube, which we can organize by conjugacy class:
the identity element fixes all $4^6$ labelings
six $90^\circ$ face rotations each fix $0$ labelings
three $180^\circ$ face rotations each fix $4^4$ labelings
eight $120^\circ$ vertex rotations each fix $4^2$ labelings
six $180^\circ$ edge rotations each fix $4^3$ labelings.
This looks very similar to the calculation when we are coloring the faces from a palette with $4$ colors explained in great detail here, with one important difference: whereas a $90^\circ$ face rotation allows an independent choice of any color on either of the two faces pierced by the rotation axis and any single choice of color for the other $4$ faces, hence $4^3$ possibilities, in the present problem none of the labelings are fixed by a $90^\circ$ rotation.
Therefore, the total count is $$ \frac{4^6 + 3\cdot 4^4 + 6\cdot 4^3 + 8\cdot 4^2}{24} = 224 $$ distinct labelings.