Suppose you have only two colors for each of the panels composing your $4 \times 4 $ window, how many different windows can you create?
This is a problem from my Abstract Algebra class where right now we're seeing group actions mainly, but I don't understand the subject enough to apply it here. I had the idea that we could divide the window in a center and an exterior. The center being a smaller $2\times 2$ window and the exterior you could think of as a necklace of length $12$ as in this intuitive proof of Fermat's little theorem: https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem#Necklaces
I managed to "count" all the different possible centers, which are only 6. In the case of the string of length $12$ I'm not sure how to count them. It'd probably be easier with group actions I'm sure. Any help on any on any of these two possible paths to solve the problem would be much appreciated.
As pointed out in the comments we should define different as in two windows are different if they don't look the same under any type of rotation(s).
I will have a go at this interesting question to check if my understanding of Burnside’s Lemma is correct, following the advice from N.F. Taussig in the comments section.
For a square $4 \times 4$ window, there are four elements of the group to consider, i.e. (clockwise) rotation by 90, 180 and 270°, as well as the identity.
Neglecting symmetries, there are $2^{16}$ possible arrangements.
The identity has $2^{16}$ fixed points.
Considering the 90° rotation case, there are 16 fixed points. Indeed one need to select the colour of one of the corner cells, the colour of each of the middle cells belonging to, say, the left-most column, and the colour of the cells in the $2 \times 2$ inner window, equalling $2^4$ possibilities. The same applies for the 270° case.
A neater way to look at it, as pointed out in the comments below, is also to consider the fact that it is sufficient to select the colours in the cells of a corner quadrant, so four cells, $2^4$ possibilities.
With regards to the 180° case, for a configuration to be a fixed point the top four cells must have the same colours, in inverted order, as the bottom four cells. The same is valid for the two cells in the middle of the outmost right columns, and the two cells in the middle of the outmost left column. So there are $2^4 \times 2^2$ possibilities for the necklace (one can choose the colours of the, say, top four cells and the middle two, leftmost cells). This number is to be multiplied by 4, the number of 180°-invariant configuration for the inner $2 \times 2$ window. Then Burnside’s Lemma allows the conclusion $$ N_{4 \times 4} = \frac{2^{16} + 16 + 16 + 4 \times2^6}{4} =16456$$ distinct configuration up to rotational symmetry , see also link to the related entry in Sloane's Encyclopedia of Sequences, courtesy of Gandalf61 in the comments below.
As a check, one can apply the Lemma to the $2 \times 2$ case, for which a brute-force count is manageable:
$$ N_{2 \times 2} = \frac{2^{4} + 2 + 2 + 4 }{4} =6$$ as expected.