Linear Algebra - Conformal mapping

Question

I need to prove the following statement:

Let $V$ and $W$ be two $n$-dimensional vector space and $f:V\to W$ an isomorphism. Suppose that $g$ and $h$ are two inner products in $V$ and $W$ respectively. If $h\big(f(x),f(x)\big)=h\big(f(y),f(y)\big)$ for all $x,y\in V$ s.t. $g(x,x)=g(y,y)$ then there exists $k>0$ such that $h\big(f(x),f(y)\big)=kg(x,y)$ for all $x,y\in V$.

Let $\{e_\alpha\}$ be an orthonormal basis of $V$ and $A\in GL(n,\mathbb R)$ be the representation of $f$ respect $\{e_\alpha\}$. I only can show that all the eigenvalues of $A$ must has the same module, that is: $$\big\{\lambda\in \mathbb C:\lambda\text{ is an eigenvalue of }A\big\}=\big\{\lambda\in\mathbb C:|\lambda|=r\big\}$$ for some $r>0$. Any idea?

Answer 1

Hint: what's given is that the unit circle (all points with $g(x, x) = 1$) in $V$ maps to some circle in $W$ (all points with $h(u,u) = r$ for some $r$).

Suppose that $v$ is any vector in $V$. Then $v = \alpha x$ where $x$ is a unit vector, i.e., $g(x, x) = 1$. What's that say about the length of $f(v)$ in terms of the length of $f(x)$ (which is just $r$)?