It is true the following statement?
Let $E^n$ be an $n$-dimensional vector-space and $\langle\cdot,\cdot\rangle$ a dot product on $E^n$. If $f:E^n\to E^n$ is an isomorphism with the following property: $$\forall x,y\in > E:||x||=||y||\Rightarrow ||f(x)||=||f(y)||$$ then there exists $\kappa>0$ such that $\langle f(x),f(y)\rangle=\kappa \langle x,y\rangle$ for any $x,y\in E^n$.
For $x \in E^n$ with $||x||=1$ let $g(x):=||f(x)||$.
Since $\forall x,y\in E^n:||x||=||y||\Rightarrow ||f(x)||=||f(y)||$, $g$ is constant. Hence there is $c \in \mathbb R$ with $c \ne 0$ and
$||f(x)||=c$ for all $x \in E^n$ with $||x||=1$.
This gives: $||f(x)||=c||x||$ for all $x \in E^n$.
Its your turn to show that $\langle f(x),f(y)\rangle=\kappa \langle x,y\rangle$ for all $x,y \in E^n$ holds with $ \kappa=c^2$.