The system is still spherically symmetric about the position of the hydrogen nucleus. The wave functions will still be the same as they are, centered on the hydrogen nucleus. The only difference is that this won't be the origin in the new coordinate system. If the new originhydrogen atom is at position $\vec{r}_0$, then in the hydrogen nucleus is at $\vec{r} = - \vec{r}_0$new coordinate system, and sothen the eigenstates of the Hamiltonian will just be given by $\psi_{nlm}(\vec{r} + \vec{r}_0)$$\psi_{nlm}(\vec{r} - \vec{r}_0)$. These are exactly the same hydrogen orbitals, centered on the same point in space. We're just calling that point in space $-\vec{r}_0$$\vec{r}_0$ instead of 0.
This is equivalent to solving a particle-in-a-box system on the interval $[0,L]$ instead of $[-L/2,L/2]$. Using the latter interval, it's clear that the system has parity symmetry, and so the eigenstates will be even and odd functions (sines and cosines). If we shift the origin so that we are using the interval $[0,L]$ instead, we still get the same functions, just shifted so that they are centered at $L/2$ instead. The symmetry is not as obvious, but it's still true that the eigenstates are even or odd about the center of the box ($L/2$).