= Solution
Equivariance under $\kappa_x$ makes the first component odd in $x$; equivariance under $\kappa_y$ and $\kappa_z$ makes it even in $y,z$. Up to cubic order its allowed monomials are therefore $x$, $x^3$, $xy^2$, and $xz^2$. Absolute irreducibility makes the linear coefficient the same in each component. Equivariance under the cyclic permutation then determines the other two components from the first. Write its three independent cubic coefficients as $-a$, $-a-c$, and $-a+e$. With $X=x^2+y^2+z^2$, the resulting <signed cyclic three-mode amplitude equations> are
$$
\boxed{\begin{aligned}
\dot x&=x(\mu-aX-cy^2+ez^2),\\
\dot y&=y(\mu-aX-cz^2+ex^2),\\
\dot z&=z(\mu-aX-cx^2+ey^2).
\end{aligned}}
$$
There are no quadratic terms because each component has its own amplitude as an odd factor and is even in the other amplitudes. Higher-order equivariant terms begin at degree five; the remaining calculation concerns this displayed cubic truncation.
For reference, with $\mu,a>0$, the coordinate-axis states are $(\pm\sqrt{\mu/a},0,0)$ and their cyclic copies. The equal-magnitude diagonal states have
$$
x^2=y^2=z^2=\frac{\mu}{3a+c-e},
$$
when $3a+c-e>0$. If that denominator is negative, the generic diagonal branch belongs to the opposite parameter side instead. This makes explicit the two isotropy types from part (b).
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