Here's a question related to a long-time personal research project in combinatorial geometry.
Suppose I have two lists of similar $n$-by-$n$ orthogonal matrices $P_i$ and $Q_i$, and suppose I know that there exists some particular $n$-by-$n$ orthogonal matrix $R$ such that $$R\;P_i = Q_i R \quad\text{for all }i$$ (That's a single $R$ working across all $i$.) Is there an efficient algorithm to construct $R$ from the $P$s and $Q$s?
Certainly, the relations $RP_i=Q_iR$ establish a (potentially huge!) linear system in the entries of $R$, and the orthogonality conditions on $R$ provide additional non-linear constraints on those entries. I can plod through those in specific numerical cases, but I'm looking for a higher-level solution, since I'd like to work with the $P$s, the $Q$s, and $R$ as abstractly as possible.
For a little context: Geometrically for, say, $n=3$, the $P$s and $Q$s represent isometries (rotations, reflections, roto-reflections) of two (origin-centered) polyhedra, $A$ and $B$, that have the same isometry group but are oriented differently from one another in space. (For simplicity here, we could assume that $A$ and $B$ are congruent.) So, $P_1$ might be a $3$-fold rotation for polyhedron $A$, and then $Q_1$ would be the corresponding $3$-fold rotation for polyhedron $B$, but the axes of rotation for the two figures might not align. The target matrix $R$ effectively allows me to re-orient one polyhedron so that its isometries use the same axes of rotation (and planes of reflection) as the other polyhedron. (I'll note that I can restrict my lists of $P$s and $Q$s to matrices representing generators of the isometry group in question.)
When I'm allowed to do so, I'll offer a bounty of at least 50 points. If you show me that I've missed a solution that should've been obvious, then I'll boost the bounty to 500 points as a symbolic kick to my own head.