Diagonalizable matrices of finite odd order are the identity I want to prove that if $A^n = I$ for some odd $n \geq 1$, and $A$ is diagonalizable, then $A=I$. So if $A$ is diagonalizable, there exists $PAP^{-1}=D$ and also $PA^mP^{-1}=I$. To prove $A=I$, we need $D=I$ but how does it work?

As JMoravitz noticed, it is false over the complex numbers. Let's assume then that we're talking about reals.
The polynomial $P(X) = X^n - 1$ annihilates $A$, so the eigenvalues of $A$ are roots of $P(X)$, ie are 1 (because $n$ is odd). As $A$ is diagonalizable, $D = 1$, and $A = 1$ follows.
