Uniqueness of Unitary Similarity Transform Let's say I have a Hermitian matrix $H$ that is diagonalized by a unitary matrix $U$: $D = U^{\dagger} H U$, where $D$ is diagonal. How unique is $U$? If I stick on an overall phase factor by $U \rightarrow e^{i\phi}U$, then $U$ still diagonalizes $H$. Do I have more options than that, or is $U$ uniquely defined up to a phase? 
 A: Yes. First of all, you can add any permutation to $U$. I.e. given a matrix $A$ and a unitary matrix $U$ such that $UAU^*$ is diagonal, $PU$ still diagonalises $A$ for every permutation $P$ (note that $PU$ is still unitary), since what it does is just permuting the entries of the diagonal matrix. 
Moreover, consider the case where $A$ is the identity matrix, then every unitary matrix $U$ diagonalizes $A$. This might give you a hint: If you have eigenvalue degeneracies, you can add even more matrices to your set:
Let $A\in \mathbb{C}^{4\times 4}$ be normal (i.e. diagonalizable by unitary matrices), and suppose we have
$$UAU^*=\operatorname{diag}(\lambda_1,\lambda_1,\lambda_2,\lambda_3) $$
where $\lambda_i\neq \lambda_j$ for $i\neq j$. Then, if we consider any matrix $\hat{U}:=\operatorname{diag}(\tilde{U},1,1)$ with $\tilde{U}\in U(2)$, then $\hat{U}U$ also diagonalizes $A$. In other words: Within a degenerate subspace, one can choose any unitary to diagonalize the matrix. 
In fact, this is all the freedom in $U$ you have and you can prove this (informally) in the following way: In order to diagonalise a normal matrix $A$, the unitary matrix $U$ must contain an orthonormal basis of eigenvectors of $A$. Now, what choice in the basis do you have? First, you can always multiply all vectors with the same phase (this is what you found), second, you can put the eigenvectors in any order you like (this is uniqueness up to permutation) and third, if you have degenerate eigenvalues, you can choose any orthonormal basis of the eigenspace you like (my last comment). This gives you all the possibilities there are. 
