``everybody knows'' that a rational orthogonal matrix is a signed permutation matrix, so there are exactly $2^n n!$ such matrices in $O(n).$ Now, what if we ask for the enumeration of elements of $O(n$ with rational elements, where all the denominators are equal to $q$ (where $q$ is an arbitrary integer, though if the question is much easier for $q$ prime, that's fine too...)? I would suspect this has been studied... (one can ask the same question for arbitrary algebraic groups/number fields, of course).

``everybody knows'' that an integral orthogonal matrix is a signed permutation matrix, so there are exactly $2^n n!$ such matrices in $O(n).$ Now, what if we ask for the enumeration of elements of $O(n$ with rational elements, where all the denominators are equal to $q$ (where $q$ is an arbitrary integer, though if the question is much easier for $q$ prime, that's fine too...)? I would suspect this has been studied... (one can ask the same question for arbitrary algebraic groups/number fields, of course).