Can any statement be made on the structure of a diagonal matrix after an unitary transformation? Let $U$ be unitary and $D$ be diagonal. Can anything be stated about the structure of $$U^\dagger D U$$ then?
Also, would this change for an infinite matrix, i.e. a discrete linear operator?
 A: The n×n matrices of the form U* D U for D diagonal and U unitary are exactly the Normal matrices, that is the matrices that commute with their conjugate transpose.  The extension to hilbert-space operators are called normal operators.
If by dagger you mean the plain transpose, then I believe Horn–Johnson's book discusses these matrices, but they are less common.
A: If the entries of $D$ are real, there is a beautiful answer due to Schur. Let the entries of $D$ be $(d_1, d_2, \ldots, d_n)$. Then the possible diagonal entries of $U^* D U$ are the points in the convex hull of the $n!$ permutations of the $d_i$.
Another way to phrase this is the following: Let $d_1 \geq d_2 \geq \cdots \geq d_n$. Then $(c_1, c_2, \ldots, c_n)$ are possible diagonal entries for $U^* D U$ if and only if
$$\sum c_i = \sum d_i$$
and, for any $i_1 < i_2 < \cdots < i_k$, we have
$$c_{i_1} + c_{i_2} + \cdots + c_{i_k} \leq d_1 + d_2 + \cdots + d_k.$$
See section 5 of Bhatia's article Linear Algebra to Quantum Cohomology: The Story of Alfred Horn's Conjectures for a proof.

If $D$ is complex, $D=A+Bi$, then $\mathrm{diag}(U^* D U^*) = \mathrm{diag}(U^* A U) + \mathrm{diag}(U^* B U^*) i$. So the real and imaginary parts of the diagonal separately obey Schur's bounds with respect to $A$ and $B$. I'm not sure if you can do better than this.
