2020 AMC 12B Problem 20

For a fixed first cube, the number of second cubes matching it (up to rotation) equals the size of its rotation orbit. So the desired probability is $\tfrac{1}{64^2}\sum_{\mathcal O} |\mathcal O|^2.$ Grouping by black-face count, the orbit sizes are: $0$ or $6$ black $\to 1;$ $1$ or $5$ black $\to 6;$ $2$ or $4$ black $\to 3$ (opposite) and $12$ (adjacent); $3$ black $\to 8$ (corner) and $12$ (band). Then $$\begin{gathered} \sum_{\mathcal O} |\mathcal O|^2 \\ {}= 1 + 36 + (9 + 144) \\ \quad {}+ (64 + 144) + (9 + 144) \\ \quad {}+ 36 + 1 = 588. \end{gathered}$$ The probability is $\tfrac{588}{4096} = \tfrac{147}{1024}.$ Thus, the correct answer is D.