Let $\mathcal{C}$ denote the set of colorings of an $8\times 8$ chessboard, where each square is coloredeither black or white. Let $\sim$ denote the equivalence relation on $\mathcal{C}$ defined as follows:
two colorings are equivalent if and only if one of them can be obtained from the otherby a rotation of the chessboard. Then what is the cardinality of the set ${\mathcal{C}}/_{\sim}$ of equivalence classes of elements of $\mathcal{C}$ under $\sim$?
My attempt: There are total $2^{64}$ possible ways of coloring the whole board. Let's think of the board as $4$ quadrants, each consisting of $4\times 4$ squares, denoted by $q_1, q_2, q_3, q_4$. Now $q_1 \sim q_2 \wedge q_3 \sim q_4$ reduces $2^{32}$ possibilities. Similarly each of $q_2 \sim q_3 \wedge q_4 \sim q_1$,$q_1 \sim q_3 \wedge q_2 \sim q_4$ reduces $2^{32}$ possibilities. Also $q_1 \sim q_2 \sim q_3 \sim q_4$ adds $2^{16}$ possibilities. So I am getting $2^{64} - 3\times 2^{32} + 2^{16}$ as my answer( which is wrong). I feel this approach is both complicated and incomplete. I want to know how to approach this question.