# Can any "explicit time-dependence" of an observable in QM be also seen as a unitary transformation

If I can naturally argue the unitary time-evolution for any observable (that would be, the answer to my linked question would be "yes"), then that argument could also apply to operators that can be seen as "functions" of other operators, and hence have something like an "explicit time dependence" (I use quotations because I don't like the terminology). For example: If there was an answer to the linked question, that would mean that any operator (observable) with time dependency in the schroedinger picture could be unitarily transformed to a picture where it doesn't have time dependency at all, or to a picture where it's time dependency is completely described by a unitary time-evolution. So hence the question:

Can I express any time-dependence $$\hat{A}(t)$$ as a unitary Transformation $$\hat{A}(t) = \hat{U}^{-1}\hat{A}(0) \hat{U}$$ ?

My initial answer to the question would be that this is possible only cases where either the eigenvalues of $$\hat{A}$$ don't change (because then you easily can give a unitary transformation connecting the old and the new eigenvectors), or if the Eigenvalues are all shifted by the same amount, and I can find a conjugate operator to $$\hat{A}$$.

If the answer to this question is no in general, it would still be interesting to think about for which classes of observables (or more general, which classes of operators) the answer is yes.