When does multiplication by an orthogonal matrix preserves the eigenvalues? Let $A$ be a real $n \times n$ matrix, with rank $\ge n-1$. 

Suppose that the eigenvalues (counted with multiplicities) of $A$ are the same as the eigenvalues of $QA$ for some orthogonal matrix $Q$. Must $Q$ be diagonal?

The condition $\text{rank}(A)\ge n-1$ is necessary: If we allow $\text{rank}(A)< n-1$, then one can take $A$ to be block diagonal with the $2 \times 2$ zero matrix as its first block. Then the entire $\text{O}(2) \times \text{Id}_{n-2}$ preserves the eigenvalues.

No. Consider
$$
R=\pmatrix{0&1\\ 1&0},\ Q=\pmatrix{R\\ &R},\ A=\pmatrix{R\\ &I},\ QA=\pmatrix{I\\ &R}.
$$
