Does $[L_z,H] = 0$ imply the state is also an Eigenstate of $H$ is also an eigenstate of $L_z$? Given that the Hamiltonian $\mathcal{H}$ is rotationally invariant then we know $[L_z,\mathcal{H}] = 0$. Does that imply that an eigenstate of H is also an eigenstate of $\mathcal{H}$?
More specifically I was thinking of the problem 6015 in Lim's Problems and Solutions on Quantum Mechanics: 
A brief description of the problem and solution: 
Question:
In a scattering processes, we have asymptotically $\Psi_{final} = e^{ikz}+f(\theta,\phi)\frac{e^{ikr}}{r}$. We need to argue that $f(\theta,\phi)$ doesn't depend on $\phi$ if Hamiltonian is rotationally invariant. 
Solution:Rotational invariance means $[L_z, \mathcal{H}] = 0$. The incident state in the scattering process is $\Psi_{inital} = e^{ikz}$, which is an eigenstate of $L_z$ with eigenvalue $0$. Since angular momentum is conserved and outgoing state $\Psi_{final}$ is also an "eigenstate  of $L_z$" with eignevalue 0. Hence we conclude that $f(\theta,\phi)$ doesn't depend on $\phi$. 
Doubt here: We know if $[L_z, \mathcal{H}] = 0$, then expectation of $L_z$ doesn't change but how do we conclude that final state is also an eigenstate of $L_z$ ?
 A: Hint: "We know if $[L_z,H]=0$, then expectation of $L_z$ doesn't change...", "The incident state in the scattering process is $\psi_{\mathrm{initial}}=e^{ikz}$, which is an eigenstate of $L_z$..."
Put another way: if $L_z|\psi(0)\rangle=m|\psi(0)\rangle$ and $[L_z,H]=0$ find $L_z|\psi(t)\rangle$.
New hint: What is $|\psi(t)\rangle$ in terms of $|\psi(0)\rangle$, $H$, and $t$? What is $\left[L_z,e^{-iHt/\hbar}\right]$? Use these to compute $L_z|\psi(t)\rangle$.
Response to new edit: "Does that imply that an eigenstate of H is also an eigenstate of $\mathcal{H}$?"
First, there's still confusion about $H$ vs $\mathcal{H}$. Did you mean, "Does that imply that an eigenstate of H is also an eigenstate of $L_z$"? If so, then the answer is no, not necessarily. You have in your example $\Psi_{\mathrm{initial}}=e^{ikz}$ which is an eigenstate of $L_z$, but not $L_x$ or $L_y$. It is also not an eigenstate of $L^2$. So, a simple counterexample is $\Psi_{\mathrm{evil}} = \frac{1}{\sqrt{2}} \left(e^{ikz} + e^{ikx}\right)$. $\Psi_{\mathrm{evil}}$ is an eigenstate of $H=\frac{p^2}{2m}$ with eigenvalue $\hbar^2 k^2/2m$, but is neither an eigenstate of $L_z$ nor of $L_x$.
Similarly, $\Psi_{\mathrm{evil}}$ is also not an eigenstate of $\vec{p}$ in spite of the fact that it is an eigenstate of $p^2$.
