linear algebra - $AU = ULambda$ - single solution (by eigenvalue decomposition)?

Let $A$ be a symmetric, real matrix.

We want to find the matrices $U$ and $\Lambda$ such that $AU = U\Lambda$.

Obviously, a solution is given by the eigenvalue decomposition, where $\Lambda$ is diagonal. But is there any other solution?

In other words: if $AU = U\Lambda$, where $A$ is a known real symmetric matrix, then must $\Lambda$ be diagonal?

Observation: In my context, I also know that $\Lambda$ is symmetric. Maybe it helps.

Answer

Definitely No, In fact if $A=U\Lambda U^\dagger$ is an answer then so is
$$A=V\Lambda^\prime V^\dagger$$
For any unitary $V$ and $\Lambda^\prime=V^\dagger U\Lambda U^\dagger V$.

An extreme example is
$$A=1\Lambda 1$$