Is rotation in $\mathbb{R}^d$ unique?

Question

Let $\boldsymbol{u} \in \mathbb{R}^d$ such that $||\boldsymbol{u}||_2 = 1$ be a directional vector. Let $Q_{\boldsymbol{u}} \in \mathbb{R}^{d \times d}$ be an orthogonal matrix such that $Q_{\boldsymbol{u}} \boldsymbol{u} = \boldsymbol{e}$, where $\boldsymbol{e} = \frac{1}{\sqrt{d}} (1, \ldots, 1)$. Let $A := \{ \boldsymbol{x} \in \mathbb{R}^d: Q_{\boldsymbol{u}} \boldsymbol{x} \geq 0 \}$ be a set obtained by a rotation of the non-negative orthant of $\mathbb{R}^d$ in the direction $\boldsymbol{u}$. Does it hold that if $Q'_{\boldsymbol{u}}$ with $Q'_{\boldsymbol{u}} \boldsymbol{u} = \boldsymbol{e}$ is another (orthogonal) rotation matrix, then $\{ \boldsymbol{x} \in \mathbb{R}^d: Q'_{\boldsymbol{u}} \boldsymbol{x} \geq 0 \} = A$?

Answer 1

No, this is not true. To see this just note that in $\mathbb{R}^3$ the set $\{x\ge 0\}$ is not invariant with respect to rotations about the axis $e$ (and compose your $Q_u$ with any such rotation which does not map the coordinate axes onto each each other).

To see that my first claim is true simply look at the circle through the points $(1,0,0), (0,1,0) $ and $(0,0,1) $ and verify it's not contained in $\{x\ge 0\}$.

Answer 2

Try $d=3$ with $\bf u = \bf e$. $Q_u$ could be the identity, and $Q'_u$ could be a rotation around $\bf e$...