Do eigenfunctions of the position and momentum operators vary from one problem to another? Now the eigenfunctions of the Hamiltonian clearly differ from one problem to another since the Hamiltonian depends on the potential and hence for a different potential we get a different eigenvalue equation for the Hamiltonian hence the eigenfunctions are different each time. 
However, the eigenvalue equation for the position and momentum operators don't change (since they don't depends on the potential), so is it always the case that the eigenfunctions of momentum and position operators (which are continuous) are always the same? My gut tells me that no since the eigenfunctions must also satisfy boundary conditions which differ from one problem to another, but Griffiths book only solved this one case for the eigenfunctions of position and momentum operators and later used them as a standard eigenbasis of position/momentum for all types of problems. Even in the end of chapter problems he never asks us to solve the eigenvalue equation for the position and momentum operators for, say, the infinite well. Which left me with the aformentioned question: Are they always the same?
 A: Yes they do vary. Say you're doing quantum mechanics on a circle. $\hat p\equiv -i\hbar \partial_x$ where $x\sim x+2\pi R$ is the coordinate along the circle, will then have discrete eigenvalues $k$ and complex exponential eigenfunctions: we have
$$
\hat p e^{ikx/\hbar}=k e^{ikx/\hbar}\,, 
$$
but $e^{ikx/\hbar}$ will be a function on the circle iff it is periodic. Therefore
$$
k\frac{R}{\hbar}\in\mathbb Z\,.
$$
In contrast if you are considering quantum mechanics on say $\mathbb R^3$ then $\hat p$ will have continuous spectrum, so its "eigenvalues" $k\in\mathbb R^{3}$.
A: The determination of eigenfunctions themselves do not depend on the boundary conditions. As you have said, the operators "don't change" from problem to problem, so the eigenfunctions will not change either. Where boundary conditions come into play is whether or not a quantum system can be in a specific eigenstate of some operator. If the eigenstate is "not allowed" due to inconsistent boundary conditions the system must have (or perhaps some other reason), then it stands to reason that the system will not be found to be in this state. 
This is precisely what happens when solving for energy eigenstates of the electron in the Hydrogen atom.$^*$ We restrict the possible eigenfunctions to be periodic in the position basis (for the appropriate spatial coordinates). The eigenfunctions themselves aren't determined by periodic boundary conditions, but the possible eigenstates we can find the system in are determined by the boundary conditions. The "thrown out" eigenfunctions are still eigenfunctions of the energy operator, it's just that for this specific system we know we cannot use them.
In a more general sense, the reason why eigenfunctions are not determined by the boundary conditions is because the "eigenfunction equation" and the boundary conditions are separate things. You can write out general solutions to differential equations without specifying any boundary conditions. When you apply your boundary conditions to pick out the solutions that work for the specific system, the invalid solutions are still solutions to the "eigenfunction equation" even if you are not using them.

$^*$I know we recognized that these energy eigenstates "change" depending on the problem, but it still holds for an example of how the boundary conditions determine allowed eigenfunctions.
