Are stationary states only eigenstates of $H$? are stationary states only eigenstates of $H$? If I have an hermitian operator $O$ that commutes with $H$ so that is a constant of motion, are the eigenstates of $O$ also stationary states since a measure of $O$ over an $O$ eigenstate always gives the same eigenvalue over time?
 A: Every vector $\psi$ is an  eigenvector of $O=I$ and $I$ commutes with $H$. However it is evidently false that every vector $\psi$ is a stationary state.
A: No.  $O$ and $H$ have a common set of eigenstates but it is possible to have an arbitrary eigenstate of $O$ that is not an eigenstate of $H$.
For instance, in the hydrogen atom problem, there are eigenstates of $L^2$ which are not eigenstates of $H$: any two states with different energies but same value of $\ell$ would be an example of that, say $a\vert 100\rangle+b\vert 200\rangle$.
Such an eigenstate of $O$ would not be stationary because it is not an eigenstate of $H$.
A: An eigenstate of a conserved observable is not necessarily a stationary state.
The impulse behind your question is well-motivated, one could think as follows:

If an observable $O$ is a conserved quantity and the system at a time $t=0$ has a definite value $o$ for the observable $O$ (i.e., the state of the system $\vert\psi\rangle_{t=0}$ is an eigenstate $\vert o\rangle $ of the observable $O$) then the value of this quantity cannot change over time (because of the observable is conserved) and thus, the state of the system at some time $t=t$ should also be the same eigenstate $\vert o\rangle$, i.e., $\vert \psi\rangle_{t=t}=\vert o\rangle$ up to an overall phase. In other words, we have shown that an eigenstate of a conserved observable is a stationary state.

However, this is incorrect. The reason is degeneracy. In particular, the following assumption that I made in the argument I gave above is not necessarily true:

If a system has a definite value $o$ for an observable $O$ then the eigenvalue $o$ is enough to label the state of the system $\vert \psi\rangle$, i.e., you can write $\vert \psi\rangle=\vert o\rangle$ with the understanding that $o$ is an eigenvalue of the observable $O$.

You can do this only when the observable $O$ is such that its eigenspaces are non-degenerate. If they are degenerate then there exist physically distinct states $\vert\psi_1\rangle,\vert\psi_2\rangle$ such that $O\vert\psi_1\rangle=o\vert\psi_1\rangle$ and $O\vert\psi_2\rangle=o\vert\psi_2\rangle$ but $\vert \langle\psi_2\vert\psi_1\rangle\vert\neq 1$. In other words, the use of the eigenvalue $o$ as a label is not enough to specify a unique state. You further need the eigenvalues of some other observable or observables that lift the degeneracy, i.e., you need further quantum numbers. And this is precisely the reason why $O$ can be conserved and yet a system prepared in an eigenstate of $O$ need not be stationary. You see, in my example with $\vert\psi_1\rangle$ and $\vert\psi_2\rangle$, a system prepared in any linear combination of those two states would be an eigenstate of $O$ with eigenvalue $o$ and as long as the time evolution keeps the system in the vector-space spanned by $\vert\psi_1\rangle$ and $\vert\psi_2\rangle$, the value of $O$ will not change. Thus, the time-evolution can evolve the state of the system to any of the infinitely many physically distinct states in the vector-space spanned by $\vert\psi_1\rangle$ and $\vert\psi_2\rangle$ without changing the value of the observable $O$.
So, the conservation of $O$ does not require the system to be stationary, it only requires the evolution to be such that it remains in the same eigen-subspace of $O$ that it starts with. This is exactly what is ensured by its commutation with the Hamiltonian.
A: The Coulomb potential (the simplest model of the hydrogen atom) is spherically symmetric, so the angular momentum operator commutes with the Hamiltonian (therefore angular momentum is conserved), so we can choose a basis of (stationary) energy eigenstates which are also eigenstates of the angular momentum operator.
