 Solution

For a uniform classical spin $S\mathbf{n}$ with ${\mathbf{n}}^2=1$, the energy per bond isThe strict inequalities among the couplings make the $x$ direction the unique easy axis. ThusThese uniform configurations are also global classical minima: the largest singular value of the diagonal exchange matrix is $J^x$, so each bond has energy at least $-J^x{S}^2$, and both configurations saturate that bound.

A global rotation by $\pi$ about the $z$ axis sends $(S^x,S^y,S^z)$ to $(-S^x,-S^y,S^z)$, preserves every exchange product, and exchanges the two ground states. A $\pi$ rotation about $y$ does the same. The broken discrete symmetry can be represented by either operation.