= Solution
Use the natural <dihedral group> generators $\rho z=e^{2\pi i/n}z$ and $mz=\bar z$, and orient the unfolding parameter so the trivial <equilibrium> loses stability as $\mu$ increases through zero. Rotation equivariance requires a <monomial> $z^p\bar z^q$ in $\dot z$ to have $p-q\equiv1\pmod n$, while reflection makes its coefficient real. These constraints give the <dihedral steady-state normal form>. The reflection axes have angles $\theta=k\pi/n$ and one-dimensional <fixed-point subspaces of a group action>; their <equilibria> are the branches guaranteed by the <equivariant branching lemma>.
For $n=3$, a generic quadratic anisotropy dominates:
$$
\dot z=\mu z+b\bar z^2+a z|z|^2+\cdots,\qquad b\ne0.
$$
In <polar coordinates>, $\dot r=\mu r+br^2\cos3\theta+O(r^3)$ and $\dot\theta=-br\sin3\theta+O(r^2)$. At a nonzero <equilibrium>, $\chi=\cos3\theta=\pm1$ and $r=-\mu/(b\chi)+O(\mu^2)>0$. There are three symmetry-related <equilibria> on each parameter side, on opposite half-axes as the sign of $\mu$ changes. The radial and angular <eigenvalues> are $-\mu+O(\mu^2)$ and $3\mu+O(\mu^2)$, respectively. Hence \b[all three nonzero branches are saddles in the full plane], even though radial stability exchanges at the crossing. The origin is a sink for $\mu<0$ and a source for $\mu>0$. The reflection <normaliser> has no residual sign reversal in this odd dihedral case, so these are transcritical-like branches, not symmetry-forced pitchforks.
For $n=4$, both relevant nonlinearities are cubic:
$$
\dot z=\mu z+a z|z|^2+b\bar z^3+O(|z|^5),\qquad
\dot r=r\bigl(\mu+(a+b\cos4\theta)r^2\bigr),\quad
\dot\theta=-br^2\sin4\theta.
$$
There are two conjugacy classes of reflection isotropy: the coordinate axes, with $\chi=\cos4\theta=1$, and the diagonals, with $\chi=-1$. Generically $b\ne0$ and $a\pm b\ne0$. Each class has a <pitchfork bifurcation>, with four symmetry-related <equilibria> when its squared radius is positive:
$$
\boxed{r_\chi^2=-\frac\mu{a+b\chi},\qquad
\rho_{\mathrm{rad}}=-2\mu,\qquad
\rho_{\mathrm{ang}}=-4b\chi r_\chi^2.}
$$
These formulas give all generic existence and stability cases. If $a<-|b|$, both classes emerge for $\mu>0$; the class with $b\chi>0$ consists of stable nodes and the other of saddles. If $a>|b|$, both classes exist for $\mu<0$; the class with $b\chi>0$ consists of saddles and the other of sources. If $|a|<|b|$, one class exists on each side and both consist of saddles. Equality $a=\pm b$ is nongeneric and requires higher-order terms. Unlike the odd dihedral cases, the residual <normaliser> quotient contains a sign reversal, enforcing exact paired pitchfork branches.
For $n=5$, cubic radial saturation occurs before the first angular anisotropy:
$$
\dot z=\mu z+a z|z|^2+b\bar z^4+O(|z|^5),\qquad
\dot r=r\bigl(\mu+a r^2+b r^3\cos5\theta+\cdots\bigr),\quad
\dot\theta=-br^3\sin5\theta+\cdots.
$$
For generic $a,b\ne0$, ten <equilibria> lie on the reflection half-axes whenever $-\mu/a>0$, divided into two distinct five-point group orbits. Their radii and <eigenvalues> are
$$
r_\chi=\sqrt{-\mu/a}+O(\mu),\qquad
\rho_{\mathrm{rad}}=-2\mu+O(|\mu|^{3/2}),\qquad
\rho_{\mathrm{ang}}=-5b\chi r_\chi^3+O(r_\chi^4),\quad\chi=\pm1.
$$
If $a<0$, they emerge for $\mu>0$: the five with $b\chi>0$ are stable nodes and the other five are saddles. If $a>0$, they exist for $\mu<0$: the five with $b\chi>0$ are saddles and the others are sources. The leading amplitude scaling is pitchfork-like, but the quartic term distinguishes opposite half-axes and generally gives different corrections to their radii; there is no exact sign-reversal symmetry on an individual fixed line. Thus \b[$D_3$ has quadratic saddle branches, $D_4$ has two cubic axial branch types, and $D_5$ has cubic radial branches whose angular stability splits at quartic order].
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