How does the wavefunction inherit the symmetries from the Hamiltonian? The question is best explained with an example,
Consider a particle in a 2D Square Box. The Hamiltonian has symmetry group $ C_{V4}$.
Consider the ground state eigenfunction, $$\psi_0 = sin(x)sin(y)$$
ignoring normalization and taking appropriate length of the box.
This w.f inherits all the symmetries of the Hamiltonian.
For the first excited state, there is a twofold degeneracy, and the WFs are:
$$ \psi_{1,2} = sin(x)sin(2y)$$
$$ \psi_{2,1} = sin(2x)sin(y)$$
The individual wavefunctions themselves do not inherit all the symmetries like in the last case. But they do preserve some symmetries.
Is there a formal way to classify how the wavefunction inherits those symmetries? And do the symmetries hold when we take into account the time evolution?
 A: What you're looking for is group theory (or more precisely representation theory), and you can start by reading about [Wigner's Theorem][1]. Eigenstates of a Hamiltonian will fall under an irreducible representation (or "irrep") of the symmetry group of the Hamiltonian. The symmetry of an irrep is less than or equal to the symmetry of the Hamiltonian. An important aspect of Wigner's theorem is that it concerns eigenstates, so the symmetry of eigenstates do not change in time (i.e. they can definitively be labeled by an irrep of the symmetry group). Conversely, any state which changes symmetry in time is not an eigenstate.
You can go further and read about character tables, which have all the relevant information about (discrete) groups as well. See the table for $D_4$ [here][2] which is relevant to your stated problem which has a Hamiltonian with the symmetry of a square.
In your case, the ground state has the full symmetry of a square, $D_4$, so it falls under the $A_1$ irrep, which is non-degerenerate. On the other hand, the first excited states with $\sin(2x)$ or $\sin(2y)$ terms are doubly-degenerate and have a lower symmetry. These two states fall under the irrep $E$ in the character table. Eigenstates falling under this irrep do not have four-fold symmetry ($C_4$ rotations bring one of the two states to the other), instead they have only one $C_2$ symmetry which gives them a negative sign.
You can even make more interesting observations just from the character table of your symmetry group. For example, there will not be any eigenstates in your problem which are more than two-fold degenerate. This follows from the fact that there are no triply-degenerate (or higher) irreps in $D_4$.
Finally, it is worth noting that you can add spin to the problem as well by considering so-called "double-groups". These are spatial symmetry groups combined with the $SU(2)$ symmmetry group of the spin operator. So the framework of representation theory applies for any symmetry, spatial or not.
And in case you want to study continuous symmetries, rather than discrete ones, then you can read up on Lie groups which are a lot nicer than finite groups in many ways.
[1]: https://en.wikipedia.org/wiki/Wigner%27s_theorem
[2]: http://gernot-katzers-spice-pages.com/character_tables/D4.html
