Can we obtain $Q$ in $$R_2 = Q^TR_1Q$$ where $R_1, R_2, Q$ are all 3x3 rotation matrices, $R_1 \ne R2$, and $\operatorname{trace}(R_1)=\operatorname{trace}(R_2)$?
I have a problem where $R_1$ and $R_2$ are known. I also have the ground truth of $Q$, which needs to be solved. The matrices do not commute since they are in 3D.
I tried vectorization in an attempt to determine $Q$ from the Kronecker root of $Q^T\otimes Q^T$,
$$\operatorname{vec}(R_2) = (Q^T\otimes Q^T)\operatorname{vec}(R_1)$$
with $\otimes$ denoting the Kronecker product. However, the approach does not seem right since it results in an underdetermined system having $\operatorname{vec}(R_2) \operatorname{vec}(R_1)^\dagger$ with a unit rank and $Q^T\otimes Q^T$ with a rank of $9$.
Is $Q$ solvable and uniquely determined, leveraging the condition that all three matrices are rotations?
I updated the title (Title was modified following a useful comment) to consider treating this as a two-sided orthogonal Procrustes problem,
$$\underset{Q}{\operatorname{argmin}}\|Q^TR_1Q - R_2\|_F^2 \ .$$
I'm also searching the literature if the problem can be solved this way. Please help.