If $\Vert \alpha(v) \Vert = \Vert v \Vert$ for all $v$ in an orthonormal basis does that mean $\alpha^{\star}\alpha = \sigma_1$? More specifically than the title, let $\alpha : V \to V$, where $V$ is a finite-dimensional inner product space. If $ \|\ \alpha(v) |\| = \|\ v \|\ $ for all $v$ in some orthonormal basis for $V$, must $\alpha$ be unitary? 
I believe $\alpha$ must be unitary with these conditions. However, I'm not 100% convinced with my proof, I feel like I'm overlooking something. 
Since $ \|\ \alpha(v) \|\ = \|\ v \|\ $, we have that for all $v \in V$,
$$ \langle \alpha(v), \alpha(v) \rangle = \langle v , v \rangle \iff \langle v , \alpha^{\star}\alpha(v) \rangle = \langle v , v \rangle $$
Since we're in an inner product space, it must be the case that,
$$\alpha^{\star}\alpha(v) = v \iff \alpha^{\star}\alpha = \sigma_{1}$$ 
and hence, $\alpha$ must be unitary. 

Consider $V = \mathbb R^2$ and $\alpha = \begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}$, take the standard basis.
