2012 AMC 12A Problem 15

The four corners form one cycle under the rotation, the four edge squares form another, and the center is fixed. These three groups are independent. A position remains white exactly when both it and the square rotated into it were originally white. Thus the corners end black exactly when their cyclic string has no adjacent pair of whites. The allowed strings are the all-black string, the $4$ strings with one white, and the $2$ strings with two opposite whites: $7$ of the $2^4=16$ possibilities. Hence the corner probability is $\frac{7}{16}.$ The same argument applies to the four edge squares. The center is black at the end only if it started black, with probability $\dfrac12.$ Multiplying, the whole grid is black with probability $$\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.$$ Thus, the correct answer is A.