Suppose A is a normal matrix over $M_n(\mathbb{C})$, with diagonalization $A = PDP^*$. Consider the inner product $<A\mathbf{v}, \mathbf{v}>$
$<A\mathbf{v}, \mathbf{v}> = <P(DP^*\mathbf{v}), \mathbf{v}> = <DP^*\mathbf{v}, P^*\mathbf{v}>$
by the properties of the adjoint of $P$. Now, since $P, P^*$ are unitary, I know that $<P^*\mathbf{v}, P^*\mathbf{v}> = <\mathbf{v}, \mathbf{v}>$. Can I therefore conclude that
$<DP^*\mathbf{v}, P^*\mathbf{v}> = <D\mathbf{v}, \mathbf{v}>$?
No, this doesn't work. For a simple counterexample, take $n=2$, $D=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$, $P=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$, and $v=(1,0)$. Then $\langle Dv,v\rangle=1$ but $\langle DP^*v,P^*v\rangle=0$.
Really, you shouldn't expect this to be true, because if you write $u=P^*v$, all you know is that $u$ is some vector with the same norm as $v$. So you should not expect that $\langle Du,u\rangle=\langle Dv,v\rangle$; there is no reason $\langle Dv,v\rangle$ should depend only on the norm of $v$.