We have $5 \times 5$ chessboard floating in space , so rotation and reflection are allowed. We want to color it using $m$ distinct colors such that when we color a square , the both sides of that square will be colored , so the both faces of chess board will be coloring while we are coloring one face. (You may think that this chessboard is transparent ) How many nonequivalent coloring are there ?
My work:
$\psi(\pi_0^*)=m^{25}$ , the identity permutation.
$\psi(\pi_1^*)=m^{7}$ , $90$ degree rotation.
$\psi(\pi_2^*)=m^{13}$ , $180$ degree rotation.
$\psi(\pi_3^*)=m^{7}$ , $270$ degree rotation.
$\psi(r_1^*)=m^{15}$ , horizontal reflection.
$\psi(r_2^*)=m^{15}$ , vertical reflection.
$\psi(r_3^*)=m^{15}$ , diagonal reflection (positive slope diagonal).
$\psi(r_4^*)=m^{15}$ , diagonal reflection (negative slope diagonal).
$$\frac{1}{4}(m^{25}+2m^7 +m^{13} +4m^{15})$$
Is my solution correct ?
EDITED ANSWER: $$\frac{1}{8}(m^{25}+2m^7 +m^{13} +4m^{15})$$
You are very close. Remember, there are $8$ possible transformations, so you need $\frac18$ in front, not $\frac14$. (As a quick check, what should the answer be if there is only a single color available, i.e. $m = 1$?)
Apart from that, I agree with all your calculations, and also with your decision to go with Burnside's lemma.