= Solution
For the <Equivariant branching lemma>, assume a generic steady-state crossing $DF(0,\mu)=\mu I$ after parameter normalization, and nonzero leading nonlinear coefficient on each axial line. The lemma gives an <equilibrium> branch for each <axial subgroup> with a one-dimensional real <fixed-point subspace of a group action>.
For $H_{\mathrm{ax}}=\langle\kappa_y,\kappa_z\rangle$, $\operatorname{Fix}(H_{\mathrm{ax}})=\{(r,0,0)\}$. Conjugation by $\rho$ supplies the other two coordinate axes. For $H_{\mathrm{diag}}=\langle\rho\rangle$, $\operatorname{Fix}(H_{\mathrm{diag}})=\{(r,r,r)\}$. Sign reflections supply all the equal-magnitude diagonals. These isotropy types are not conjugate: their subgroup orders are $4$ and $3$. Thus \b[there are two distinct primary axial branch types: coordinate-axis states and equal-magnitude diagonal states].
The global sign reversal belongs to the group and makes the restricted scalar equations odd, so generically each has the form $\dot r=r(\mu+\gamma r^2+\cdots)$, with $\gamma\ne0$. Its two nonzero solutions satisfy $r^2=-\mu/\gamma+\cdots$ on the appropriate parameter side. Symmetry-related solutions belong to the same branch type; their existence does not imply stability or that both types lie on the same side of the crossing.
Back to article page
