Orthogonal matrix $Q$ such that $\forall x\leq 0$, $Qx\geq 0$

Question

What are the orthogonal matrices $Q$ such that for all vectors $x\leq 0$, $Qx\geq 0$? The inequality is to be understood component-wise.

In dimension 1, the only possibility is $Q=[-1]$, which is a reflection.

In dimension 2, $Q$ maps the third quadrant to the first quadrant (and vice-versa). It could be the rotation of angle $\pi$ or the reflection with respect to the second bissector:

$$Q=\begin{bmatrix} -1& 0 \\ 0 & -1 \end{bmatrix} \quad \text{or}\quad \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$$

I can't think of other such orthogonal matrix (but that's not a proof...).

What about the general case (finite dimension)?

Answer 1

By computing $Q(-e_i)$ for all $i$ you can see that all of $Q$'s entries must be nonpositive.

This means that two columns of $Q$ are orthogonal only if they are structurally orthogonal; since all columns must have at least one nonzero entry, this implies that all of the columns have exactly one nonzero entry.

Therefore the only $Q$ satisifying your condition are matrices $-P$ where $P$ is a permutation matrix.