= Solution
Seek an <equilibrium> in the invariant coordinate plane $z=0$ with $x,y\ne0$. Put $U=x^2$ and $W=y^2$. The two nonzero-component equations are
$$
\mu=a(U+W)+cW,\qquad \mu=a(U+W)-eU.
$$
Subtracting gives $eU+cW=0$, so positive $U,W$ require $ce<0$. With $D=a(c-e)-ce$, the solutions are
$$
\boxed{x^2=\frac{\mu c}{D},\qquad y^2=-\frac{\mu e}{D},\qquad z=0.}
$$
They are actual additional <equilibria> when $D$ has the sign of $c$; the signs of $x,y$ are independently free, and cyclic permutation supplies the other two coordinate planes. For example, $a=\mu=c=1$, $e=-1$ gives $D=3$ and $x^2=y^2=1/3$. Thus \b[additional two-coordinate <equilibria> do exist for suitable opposite-sign cross-couplings]. The inequality $ce<0$ is necessary for this type, but without the positivity conditions on the displayed denominator it need not be sufficient for every coefficient choice.
The isotropy of a generic such point is just $\langle\kappa_z\rangle$, whose fixed-point subspace is two-dimensional. The <Equivariant branching lemma> guarantees axial branches with one-dimensional fixed spaces; it does not rule out these lower-symmetry branches. Their existence therefore gives no contradiction.
At fixed $\mu>0$, letting $e$ cross zero with $c\ne0$ sends $x^2\to\mu/a$, $y^2\to0$: the new pair of opposite-$y$ branches meets an axis <equilibrium>. Its $y$ <eigenvalue> is $e\mu/a$, so this is a secondary <symmetry-forced pitchfork bifurcation>. Similarly, as $c$ crosses zero with $e\ne0$, the branches meet the $y$ axis, whose $x$ <eigenvalue> is $-c\mu/a$. The cubic formula gives $y^2\sim-\mu e/(ac)$ near the first crossing, and $x^2\sim-\mu c/(ae)$ near the second. Reflections force the two signs of the newly nonzero coordinate, while cyclic symmetry gives the corresponding pitchforks on the other axes. The simultaneous $c=e=0$ case is degenerate and is not a generic single-parameter pitchfork.
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