= Solution
Use a real physical field, so the three positive <wavevectors> are accompanied by their complex-conjugate modes. Modulo lattice translations, the effective <symmetry group> is $G=\mathbb T^2\rtimes C_6$. Its translation <group action> is
$$
T_{\theta_1,\theta_2}(A,B,C)=(e^{i\theta_1}A,e^{i\theta_2}B,e^{-i(\theta_1+\theta_2)}C).
$$
A $120^\circ$ rotation cyclically permutes the amplitudes; a half-turn $\kappa$ acts by simultaneous <complex conjugation>. Together these generate the sixfold rotation group. These formulas specify the <group action> independently of the choice of orientation of the three <wavevectors>.
For a nonzero real hexagon $(h,h,h)$, its <stabilizer subgroup> is the rotation group $C_6$. A translation fixes it only when all three phase factors equal one. Cyclic permutation forces a fixed vector to have equal components, and the half-turn forces those components to be real. Therefore $\operatorname{Fix}(C_6)=\{(a,a,a):a\in\mathbb R\}$. For a roll $(r,0,0)$ with nonzero real $r$, the <stabilizer subgroup> consists of the translations $T_{0,\theta}$ and the half-turn, with $\kappa T_{0,\theta}\kappa=T_{0,-\theta}$. It is $S^1\rtimes C_2$, isomorphic to $O(2)$ as an abstract group even though spatial reflections are absent. The circle eliminates $B,C$ from its <fixed-point subspace of a group action>, and the half-turn makes $A$ real. Thus
$$
\boxed{\operatorname{Fix}(H_{\rm hex})=\mathbb R(1,1,1),\qquad
\operatorname{Fix}(H_{\rm roll})=\mathbb R(1,0,0).}
$$
If unreduced physical translations are used, the full lattice kernel is included in both <stabilizer subgroups>. The real representation on the six critical components is absolutely irreducible: translations distinguish the three Fourier pairs, rotations permute them, and the half-turn rules out a complex scalar in the commutant. The <equivariant branching lemma> consequently supplies generic steady branches of both axial types at a transverse crossing of the common linear <eigenvalue>. It guarantees existence, not stability or exhaustiveness of the possible branches.
Translation equivariance severely restricts the monomials in $\dot A$. Through cubic order they are $A$, $\overline B\,\overline C$, $|A|^2A$, $|B|^2A$ and $|C|^2A$: each has the same translation weight as $A$, using $\mathbf k_1+\mathbf k_2+\mathbf k_3=0$. The half-turn forces their coefficients to be real; the $120^\circ$ rotation gives the two cyclic counterpart equations. A spatial reflection would interchange $B,C$ and require equal cross-couplings, but it is not a symmetry here. After a real amplitude rescaling makes the generic nonzero quadratic coefficient one, the <rotational hexagon amplitude equations> are
$$
\dot A=\mu A+\overline B\,\overline C-\nu_1|A|^2A-(\nu_2+\delta)|B|^2A-(\nu_2-\delta)|C|^2A,
$$
with the other two equations obtained by cyclic permutation. All four coefficients are real; $\delta$ records the absence of spatial reflection. The normalization assumes the resonant quadratic coefficient is nonzero; its vanishing would be an additional degeneracy.
For the roll, $r^2=\mu/\nu_1$. The radial <eigenvalue> is $-2\mu$ and the phase of the active mode is a neutral translation. The transverse variables $(b,\overline c)$ have matrix
$$
\begin{pmatrix}\mu-(\nu_2-\delta)r^2&r\\r&\mu-(\nu_2+\delta)r^2\end{pmatrix}.
$$
Its two real <eigenvalues>, each occurring twice in the full real linearization, are $-(\nu_2-\nu_1)r^2\pm\sqrt{\delta^2r^4+r^2}$. Define $d=\nu_2-\nu_1$ and $D_r=d^2-\delta^2$. Strict attraction modulo translation is therefore equivalent to
$$
\boxed{\nu_1>0,\quad \mu>0,\quad d>0,\quad D_r>0,\quad
\mu>\frac{\nu_1}{D_r}.}
$$
Indeed the larger transverse <eigenvalue> is negative precisely when $dr^2>\sqrt{\delta^2r^4+r^2}$, and squaring is legitimate only after requiring $d>0$. If $\delta^2>d^2$, that inequality is impossible. Equality $\delta^2=d^2$ is also insufficient because of the additional positive term $r^2$. Any nonzero roll that exists with $\nu_1<0$ already has a positive radial <eigenvalue>; the degenerate case $\nu_1=0$ has no ordinary isolated nonzero roll branch.
For hexagons put $N=\nu_1+2\nu_2$. A nonzero real amplitude $h$ satisfies $\mu=Nh^2-h$, or $h=(1\pm\sqrt{1+4N\mu})/(2N)$ when $N\ne0$. Separate real and imaginary disturbances. The real <Jacobian matrix> is cyclic, with diagonal $-h-2\nu_1h^2$ and off-diagonal entries $h-2(\nu_2\pm\delta)h^2$. Its <eigenvalues> are
$$
\lambda_s=h-2Nh^2,\qquad
\lambda_\pm=-2h+2dh^2\pm2i\sqrt3\delta h^2.
$$
The imaginary <Jacobian matrix> has every entry equal to $-h$, giving <eigenvalues> $-3h,0,0$. The two zero modes translate the hexagonal pattern. Thus the general strict stability conditions for any nonzero root $h$ are
$$
\boxed{h>0,\qquad 2Nh>1,\qquad dh<1,\qquad \mu=Nh^2-h.}
$$
These conditions state <orbital stability> transverse to translations, not asymptotic convergence to one prescribed spatial phase. Equalities require a nonlinear bifurcation calculation.
When $\nu_2>\nu_1>0$, both $d,N$ are positive. The upper positive hexagon branch is radially stable after its <saddle-node bifurcation> at $\mu=-1/(4N)$, $h=1/(2N)$; it remains stable until $h=1/d$. At this second threshold the pair $\lambda_\pm$ crosses the imaginary axis with nonzero frequency if $\delta\ne0$. The crossing is transverse, since $dh/d\mu=1/(2Nh-1)$ and the derivative of $-2h+2dh^2$ with respect to $h$ is two there. Hence, subject to the usual nonzero cubic coefficient, this is a <Hopf bifurcation>:
$$
\boxed{\mu_H=\frac{2\nu_1+\nu_2}{(\nu_2-\nu_1)^2},\qquad
\omega_H=\frac{2\sqrt3|\delta|}{(\nu_2-\nu_1)^2}.}
$$
For $\delta=0$ the same threshold has two zero real <eigenvalues>, so it is not a Hopf bifurcation.
A possible <phase portrait> when neither steady branch is stable has an attracting periodic oscillation surrounding the unstable hexagon, with the three amplitudes waxing and waning cyclically. This is realized, for example, by $\nu_1=1$, $\nu_2=2$, $\delta=2$ just above $\mu_H=4$. Rolls are unstable for every positive $\mu$ because $\delta^2=4>d^2=1$. At the hexagon Hopf point $h=1$, the real spectrum is $-9,\pm4i\sqrt3$. Evaluation of the cubic <Hopf normal form> with a unit-norm critical eigenvector gives $\operatorname{Re}G_{21}=-64/27<0$, so this example has a <supercritical Hopf bifurcation> and an attracting small <limit cycle> in the real invariant subspace. The figure shows such a cycle at $\mu=4.2$ and an approaching trajectory; the positive octant is forward invariant because an amplitude's derivative at zero is the product of the other two. This is a possible portrait, rather than a claim that every parameter set with unstable rolls and hexagons has the same attractor. Stability asserted for the periodic orbit here is within the stipulated real subspace.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-58-oscillating-hexagons.png]
{title=A possible real-amplitude phase portrait: an attracting oscillating hexagon with unstable steady hexagon and roll states}
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