Are eigenfunctions of $E$ also eigenfunctions of $p$? Given that $\hat{H}\Psi = \hat{E}\Psi$
and that $E=\frac{p^2}{2m}$
Assuming a non-relativistic, system, does this mean that any eigenfunction of Energy is also an eigenfunction of momentum? Or does it work the other way and every eigenfunction of momentum is an eigenfunction of Energy?
If not, where does this break down? It seems reasonable to me that two particles with the same Kinetic Energy and mass would have the same momentum. Even if the momentum is in different directions I think those would be degenerate states and not strictly count. Although I don't really understand why degenerate states are a problem yet.
The point is that any wavefunction of definite energy sounds like it would also have a definite momentum, or vice versa, and I can't think of a way that doesn't make sense.
 A: @mikestone's answer is perfect.
More broadly, if two operators commute and both have non-degenerate spectra, an eigenstate of one operator would be an eigenstate of the other and vice-versa. However, when the spectrum of any one of the operators is degenerate, this no longer remains the case.
Let's say $[\hat{A},\hat{B}]=0$. This translates to $\hat{A}\hat{B}=\hat{B}\hat{A}$. Now, consider an eigenstate of $\hat{A}$, say $\vert a\rangle$. Then, we can observe that $\hat{A}\hat{B}\vert a\rangle=\hat{B}\hat{A}\vert a\rangle=\hat{B}a\vert a\rangle=a\hat{B}\vert a\rangle$. Or, $\hat{A}\big(\hat B\vert a\rangle \big)=a\big(\hat B\vert a\rangle\big)$. In other words, $\hat{B}\vert a\rangle$ is an eigenstate of $\hat{A}$ with the eigenvalue $a$. But, by stipulation, $\vert a\rangle$ is an eigenstate of $\hat{A}$ with the eigenvalue $a$. This means that if $\hat{A}$ has a non-degenerate spectrum then it has to be the case that $\hat{B}\vert a\rangle=\lambda\vert a\rangle$ for some (real) scalar $\lambda$. Or, in other words, $\vert a\rangle$ is an eigenstate of $\hat{B}$.
Notice that this argument relies only on the non-degeneracy of the spectrum of $\hat{A}$ but not that of $\hat{B}$. This means that what we have found is that if two operators commute then an eigenstate of one of the operators is necessarily an eigenstate of the other operator if the other operator's spectrum is non-degenerate. And it would follow that if both of them have non-degenerate spectra then an eigenstate of one operator would be an eigenstate of the other and vice-versa.
A: An eigenfunction of $\hat p$ is an eigenfunctions of $\hat p^2$ but not necessarily the other way round.  An eigenfunction of $\hat p^2$ can be a linear combination  $\alpha |p\rangle +\beta |- p\rangle$  with $\hat p |\pm p\rangle= \pm p |\pm p\rangle$ and this is not an eigenfunction $\hat p$ unless one of $\alpha,
\beta$ is zero.
A: Since you don't specify $\hat H$, the general answer is a resounding no. If there is a potential, say:
$$ V(x) = -\frac k r $$
then the energy eigenstates are definitely not momentum eigenstates.
Any potential other than $V(x) = c$ breaks translational invariance, so that eigenstates of energy cannot have definite momentum.
Note that $ V(x) = -\frac k r $ is spherically symmetric (it's the hydrogen atom), and in that case, the solution are angular momentum eigenstates.
