T and T*T are simultaneously diagonalisable

Question

I suspect the following might be true but I can't prove it.

Suppose $T \in \text{End}(V)$ for some finite-dimensional complex inner product space $V$, such that $T^*T = TT^*$ (i.e. $T$ is normal). Prove that if

$$T^*T = UDU^*$$

for some unitary $U$ and diagonal $D$ then the columns of $U$ are eigenvectors of T.

Equivalently, prove that $U^*TU$ is diagonal.

Answer 1

No, this does not work if $T$ has multiple eigenvalues of the same modulus.

For example, let $$ T=\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix} $$ Then $T^*=-T$, $TT^*=T^*T=I$ so we can pick $U=I$. But $T$ is not diagonal.