= Solution
Choose amplitude coordinates in which a quarter-turn and a diagonal <reflection> act as $r(A,B)=(B,-A)$ and $s(A,B)=(-A,B)$. These generate the natural two-dimensional <group representation> of the <dihedral group> $D_4$. An <equivariant dynamical system> satisfies $F(gv)=gF(v)$. The two independent sign changes make $F_A$ odd in $A$ and even in $B$, and $F_B$ odd in $B$ and even in $A$. The quarter-turn equates the two linear coefficients and the two sets of cubic coefficients. No quadratic monomial is allowed. Thus the <square-symmetric cubic steady-state normal form> is
$$
\boxed{A_T=\mu A-aA^3-bAB^2+O(|(A,B)|^5),\qquad B_T=\mu B-aB^3-bA^2B+O(|(A,B)|^5).}
$$
The <bifurcation parameter> has been scaled so that the critical linear <eigenvalue> is $\mu$. Nondegeneracy at cubic order means $a\ne0$, $a+b\ne0$ and $a-b\ne0$. The <equilibrium> equations factor as $A(\mu-aA^2-bB^2)=B(\mu-aB^2-bA^2)=0$. When both components are nonzero, subtracting the two bracketed expressions gives $(a-b)(A^2-B^2)=0$. Therefore the complete leading list of small steady branches is:
* The zero state, with two linear <eigenvalues> $\mu$.
* Four axial states $(\pm\sqrt{\mu/a},0)$ and $(0,\pm\sqrt{\mu/a})$, when $\mu/a>0$. Their <stability matrix> has <eigenvalues> $-2\mu$ and $\mu(1-b/a)$.
* Four diagonal states $(\pm s,\pm s)$, with independent signs and $s^2=\mu/(a+b)>0$. Their <eigenvalues> are $-2\mu$ and $2\mu(b-a)/(a+b)$.
Each family is one orbit of the <dihedral group>. Each nonzero state retains one <reflection> fixing its amplitude line; its full spatial symmetry depends also on the chosen planforms. Higher-order terms perturb the branch amplitudes but preserve the nondegenerate local classification. For $\mu>0$, an axial branch is <asymptotically stable> precisely when $a>0$ and $b>a$, while a diagonal branch is <asymptotically stable> precisely when $a+b>0$ and $b<a$. Every cubic branch lying on $\mu<0$ has a positive radial <eigenvalue> $-2\mu$, so it is unstable. In the exceptional isotropic cubic case $a=b$, the cubic truncation instead has a circle of <equilibria>; higher-order anisotropy is then needed, which is why that case was excluded.
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