If a linear isometry is injective on a subset of $\mathbb R^n$, is it also surjective?

Question

Excuse me if I am blinded by the obvious.

Let $X$ be an arbitrary set of unit vectors in $\mathbb R^n$ and let $Q\in O(n,\mathbb R)$. If $Q$ maps $X$ into $X$, must it map $X$ onto $X$?

Intuitively, I am tempted to say that the answer is yes, but I don't see any immediate reason why it is so. It is well known that any isometric map on a compact metric space is surjective, yet our $X$ is not necessarily closed here.

Answer 1

Take $n=2$, and take for $Q$ a rotation with angle an irrational multiple of $\pi$, so that it has infinite order. Then take $X=\{Q^m x|m\in \Bbb N\}$, for $x$ some fixed unit vector. Then $Q$ maps $X$ into $X$, but is not onto, as no vector $y\in X$ is such that $Qy=x$, since $Q^{-1}x\notin X$.

Note that the action of $Q$ on $X$ can be seen as a kind of "shift operator".