We consider the following two-dimensional Hamiltonian
$H=\frac{1}{2m}(p_x^2 + p_y^2) +\lambda(x^4+y^4)$
Find the highest $n$ such that the Hamiltonian has a $Z_n$ symmetry.
My Attempt
The group $Z_n$ consists of the elements $\{0, 1, 2, ..., n-1\}$ with addition $\mod n$ as the operation.
In order for the Hamiltonian $H$ to have a $Z_n$ symmetry
$\rho(a)^{-1}H\rho(a)=H$
should be satisfied. Such that $\rho(a)$ is the translation in space due to $a\in Z_n$ and $\rho(a)^{-1}$ is its inverse. Such that
$\rho(a)|\psi\rangle = \{\psi(x-a)\}$
It is easy to see that
$\rho(a)^{-1}p_x\rho(a)=p_x$ (same for $y$ direction)
However
$\rho(a)^{-1}x^4\rho(a)|\psi\rangle = \rho(a)^{-1}\{x^4\psi(x-a)\} = (x + a)^4|\psi\rangle $
Therefore, in order for the Hamiltonian to have a $Z_n$ symmetry, $a$ should be zero meaning that $n$ should be $1$.
I must confess that my knowledge of group theory and symmetry is still very much limited, I therefore apologize if I made any stupid mistakes in my attempt.
Any help would be very much appreciated.