Suppose I have two operators, $A$ and $B$, with eigenstates $A \lvert a \rangle = a \lvert a \rangle$ and $B \lvert b \rangle = b \lvert b \rangle$, where $a$ and $b$ are all unique. Furthermore, suppose that $A$ and $B$ are related by a unitary transformation $$A = U B U^{-1}.$$ This is equivalent to saying that the eigenstates are related as $$\lvert a \rangle = U \lvert b \rangle.$$ 

Then it seems I can prove the following: since $$A \lvert a \rangle = a \lvert a \rangle,$$ I also have $$A U \lvert b \rangle = U B U^\dagger U \lvert b \rangle$$ by inserting the identity, so that $$A U \lvert b \rangle = U B \lvert b \rangle = b U \lvert b \rangle = b \lvert a \rangle.$$ Thus, $a = b$. 

Doesn't this imply then that the eigenvalues for corresponding eigenstates of $A$ and $B$ are equal, and therefore-- by the assumption that they are unique-- that the unitary transformation doesn't actually *do* anything?