What happens when two diagonal matrices are unitarily similar?

Question

I'm given two diagonal matrices $D$ and $E. $ And I've managed to show that $D=UEU^*$ where $U$ is a unitary matrix. I have to show that $D$ and $E$ are related to a permutation: $D=PEP^T$ where P is a permutation matrix. I'm getting confused because I'm getting the strong result that $d_i=e_i$ i.e. $D=E$ from computing the diagonal entries in $DU=UE$ with just these assumptions. When the problem only requires the weaker result $D=PEP^T$. I haven't even used one of the assumptions in the problem! That all the entries on D and E are complex and lie on the unit circle. I don't know what I'm doing wrong or if there's a typo on the text.

Answer 1

Since $D$ and $E$ are conjugate by $A$ (i.e., $E = ADA^{-1}$), they must have the same eigenvalues with eigenspaces of the same same dimension: if $Dx = \lambda x$, then $E(Ax) = ADx = \lambda Ax$.

It follows from this that the diagonal entries of $E$ and $D$ are the same, in a possibly different order. If $D$ has diagonal entries $d_i$, consider $\{(i,d_i): i = 1,\dots, n\}$. Since the diagonal entries $e_i$ of $E$ are the same, we can find a permutation $\sigma$ with $\{(i,e_i)\} = \{(i, d_{\sigma(i)})\}$. Your $P$ is the matrix corresponding to this $\sigma$.

Notice, this did not depend on unitary similarity.