2012 AMC 10A Problem 20

The center square must initially be black, contributing probability $\dfrac12$. The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle. For one $4$-cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are $BBBB$, the four rotations of $BBBW$, and the two rotations of $BWBW$, for $7$ colorings out of $16$. The same count applies to the edge-middle cycle, independently. Therefore the probability is $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$. Thus, A is the correct answer.