If v and w are orthogonal vectors and their images f(v), f(w) are orthogonal there exists an isometry we can build with a scalar $\alpha$

Question

Let V be an inner product space.

$f: V \to V$ and f is linear for factors in $\mathbb C$, $f \neq 0 $

Prove that if $ \langle v, w \rangle = 0 $ then $ \langle f(v), f(w) \rangle = 0 $ for all $v, w \in V$

then there exists $\alpha \in \mathbb C$ so that: $\alpha f: V\to V$ is an isometry.

Which means: $ \langle \alpha f(v), \alpha f(w) \rangle = \langle v, w \rangle $ for all $ v, w \in V$

I'd really appreciate an explanatory answer, because I have no clue at the moment and fizzled around 3 hours with this question.
Have a great start in your week.:)