Unitary equivalence of matrices with equal trace It is obvious that $B = OAO^\dagger$ have equal trace if $O$ is Hermian. But is is also so that: $Tr(A) = Tr(B) \rightarrow A = OBO^\dagger$ For some Hermitian $O$ ? I know this should be trivial to prove/disprove but Is just can't get it sorted out...

I think that you meant that $O$ is _unitary_ , and that $A$ and $B$ are Hermitian. If that's the case, your statement is not true.

Note that $OAO^\dagger$ has the same eigenvalues as $A$. If we take $$ A = \pmatrix{1&0\\\0&1}, \quad B = \pmatrix{2&0\\\0&0} $$ Then we see that $\operatorname{tr}(A) = \operatorname{tr}(B)$. Since they have different eigenvalues, however, there can be no unitary $O$ such that $B = OAO^\dagger$.