I am trying to prove the following proposition:
If $f$ is an endomorphism of an Euclidean space $(E,\langle \cdot, \cdot \rangle)$, then TFAE:
$f$ is an isometry of $E$.
The representative matrix of $f$ relative to any orthonormal basis of $E$ is orthogonal.
There exists an orthonormal basis of $E$ for which the representative matrix of $f$ relative to this basis is orthogonal.
I'm stuck on how to prove that the last implies the first.
Proof of the others:
Let $\{u_1, \ldots, u_n\}$ be an orthonormal basis of $E$. As $f$ is an isometry, then $\{f(u_i), i\}$ is an orthonormal basis as well.
Let $A = (a_{ij})$ be the representative matrix of $f$ relative to the $u_i$'s. Then, $a_{ij} = \langle f(u_j), u_i \rangle$.
Let $B = A A^t = b_{ij}$. Then:
$$b_{ij} = \sum_{t = 1}^n a_{it}a_{jt} = \sum_{t=1}^n \langle f(u_t), u_i \rangle \langle f(u_t), u_j \rangle = \langle u_i, u_j \rangle = \delta_{ij}$$
Therefore $B=I_n$ and this completes the proof.
The other one is obvious.
Thanks for your help.
Observe that if $\{e_i\}$ is an orthonormal basis, the respective Gram matrix representing $\langle \cdot, \cdot \rangle$ is $(\langle e_i, e_j \rangle)$, which is the identity matrix.
So $\forall v\in E$, $\langle fv , fv\rangle = v^tf^tfv= v^tv = \langle v , v\rangle$.