$|\langle v,w \rangle |=0 \implies \langle f(v),f(w) \rangle=0 $ then there is a constant $\lambda \in \mathbb{R}$ such that $\lambda f$ is isometric. Let $(V, s)$ be an  $n$-dimensional euclidian vector space and let $f: V \to V$ be an isomorphism. 
Show that if $|\langle v,w \rangle |=0 \implies \langle f(v),f(w) \rangle=0 $ there is a constant $\lambda \in \mathbb{R}$ such that $\lambda f$ is isometric. 
That would mean that the function maps orthogonal vectors onto orthogonal vectors. If one would consider a orthogonal basis $B$ of $V$ that would mean that there is an orthogonal basis of $V$ made of $f(B)$. It is a basis because $f$ is an isomorphism and orthogonal because of the given condition. 
However the longitude of the vectors in $f(B)$ is not guaranteed to be constant or proportional to those in $B$. Not even for an orthonormal basis $B$. Is that correct? Can I somehow normalize them to a length? If not, how would I continue? 
 A: You have gotten pretty far.
Let $v_1,\dots,v_n$ be an orthonormal basis for $V$.
You figured out that the vectors $w_i=f(v_i)$ are an orthogonal basis.
The questions is whether each $w_i$ has the same length.
Instead of considering just the basis vectors, take simple linear combinations.
The vectors $v_1+v_2$ and $v_1-v_2$ are orthogonal.
Therefore
$$
0
=
\langle f(v_1+v_2),f(v_1-v_2)\rangle
=
 \langle w_1,w_1\rangle
-\langle w_1,w_2\rangle
+\langle w_2,w_1\rangle
-\langle w_2,w_2\rangle
=
|w_1|^2-|w_2|^2.
$$
Therefore $w_1$ and $w_2$ have the same length.
You can apply the same to any pair.
Therefore there is a single number $\lambda>0$ so that $|w_i|=\lambda$ for all $i$.
Now $\lambda^{-1}f$ is an isometry, which should not be hard to prove:
it conserves the norms of the basis vectors and orthogonality, so it has to preserve all inner products.
Why should one realize to look at linear combinations, then?
You already used the basis vectors themselves.
That was not enough, so the next thing is to look at their linear combinations.
Linear combinations often give something non-trivial, since the inner product is bilinear.
Also, it is a known trick and related to the polarization identity.
A: Let $x,y\in V$ be with unit norm:
$$|x|=|y|=1$$
Then, $\langle x + y, x - y \rangle = |x|^2-|y|^2=0$, so $x+y$ and $x-y$ are orthogonal. 
Since $f$ maps orthogonal vectors to orthogonal vectors, we get
$$0=\langle f(x+y),f(x-y)\rangle=|f(x)|^2-|f(y)|^2 \Rightarrow \\
\Rightarrow |f(x)| = |f(y)|$$
I hope you can proceed from here.
