quantum mechanics - What is the symmetry group of this Hamiltonian?

Consider a Hamiltonian

$$\hat H=-\partial_x^2-\partial_y^2+(x-y)Q,$$

where $x,y\in[0,a]$ (homogeneous Dirichlet boundary conditions assumed), and $Q$ is some real parameter.

When $Q=0$, the Hamiltonian describes a 2D particle in a box, and the symmetry group is dihedral group of order 8, which has two complex conjugate irreducible representations, giving rise to a two-fold degeneracy of spectrum.

But choosing $Q\ne0$ should lift this degeneracy, because, as it seems, the symmetry reduces to cyclic group of order 2. But when I solve Schrödinger's equation with this setting, the degeneracy doesn't appear to be lifted, no matter how large $Q$ I take! So, I must be missing some other symmetry operations than just reflection with respect to $x=a-y$ line.

So, what are these extra symmetry operations? Or is the extra symmetry here something hidden like dynamical symmetry of Coulomb potential?

What's interesting, the degeneracy is lifted when I take $(x-y)^\alpha$ instead of $(x-y)$, where $\alpha\ne1$. Also what's common between the cases of $Q=0$ and $Q\ne0$ is that Schrödinger's equation is separable, unlike the case of $\alpha\ne1$.

Actually, $\hat H$ is a sum of two Hamiltonians acting on independent degrees of freedom and having identical spectra. This somewhat gives intuitive expectation that the spectrum must be degenerate: there do exist states like $\left|a\right\rangle\otimes\left|b\right\rangle$ as well as $\left|b\right\rangle\otimes\left|a\right\rangle$ (with appropriate modifications), which must have equal energies, but still: what symmetry group results to explain the degeneracy from symmetry POV?