I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!
Let $V \ne 0$ be an finite-dimensional inner complex vectorspace and $f$ an endomorphism on said space such that there exists a $m \in \mathbb{N}_{\ge 1}$ with $f^m=id_V$. Given that all eigenspaces are orthogonal, I have to show that $f$ is unitary. The converse I already have proven.
I thought about starting with $f^m=id_V$:
If we take an eigenvector $x$ to the eigenvalue $λ$ we get: $f(x)=λx$, therefore $f^m(x)=λ^mx$. But $f^m(x)=x$ since $f^m=id_V$. Together: $λ^mx=x$, so: $λ^m=1$.
It follows that all eigenvalues of $f$ are $m$-th roots of unity and then: $\bar{\lambda}=1/λ$
This is the part where I unfortunately get stuck on :P
Any Ideas?
~Cedric :)