= Solution
The axisymmetric magnetic mode spans a one-dimensional real representation of axial rotations, so its signed amplitude is a real variable $B$. An equatorial dipole has two real horizontal components. Rotation through an angle $\vartheta$ rotates that pair; encoding the components by one <complex amplitude> gives $A\mapsto e^{i\vartheta}A$, and rotation through $\pi$ gives $A\mapsto-A$. The velocity mode has the same rotational representation, so write it as $V\mapsto e^{i\vartheta}V$. Magnetic reversal acts as
$$
(B,A,V)\longmapsto(-B,-A,V).
$$
Thus the two relevant symmetries are axial rotation and magnetic reversal. Reflection symmetry has not been assumed, so complex coupling coefficients are permitted.
The equations for $B,A$ must be linear in the magnetic variables at fixed velocity. Combining this with rotational <equivariance>, their general local form is
$$
\begin{aligned}
\dot B&=Bf(|V|^2)+A^*Vg(|V|^2)+AV^*g(|V|^2)^*,\\
\dot A&=Ah(|V|^2)+BVj(|V|^2)+A^*V^2\ell(|V|^2),
\end{aligned}
$$
where $f$ is real valued and $g,h,j,\ell$ are complex valued. These functions can be understood through their local Taylor expansions; the symmetry argument does not require a convergent infinite series. Reality of $\dot B$ explains its conjugate pairing.
For the velocity equation, a <monomial> $B^pA^q(A^*)^rV^s(V^*)^t$ must have the same rotational weight as $V$ and be even under magnetic reversal. Its selection rules are
$$
q-r+s-t=1,\qquad p+q+r\text{ is even}.
$$
A formal sum of all such <monomials> with complex coefficients is the most general local velocity evolution. At degree one there is $V$; at degree two there is $AB$; at degree three there are $V|V|^2,V|A|^2,V^*A^2,VB^2$. Expanding the magnetic equations similarly gives the <equivariant three-mode dynamo normal form>
$$
\boxed{\begin{aligned}
\dot B&=\mu_1B+c_1A^*V+c_1^*AV^*-e|V|^2B,\\
\dot A&=(\mu_2+i\omega_2)A+c_2VB-d_2|V|^2A-d_3V^2A^*,\\
\dot V&=(\mu_3+i\omega_3)V+c_3AB-d_4V|V|^2-d_5V|A|^2-d_6V^*A^2-d_7VB^2.
\end{aligned}}
$$
The $\mu_i,\omega_i,e$ are real, and the remaining coefficients are complex. The minus signs on cubic coefficients are a convention, not an assumption about their signs. Terms such as $B^3$ and $A|A|^2$ are forbidden in the magnetic equations by magnetic linearity, even though symmetry alone would allow some of them.
For the <normalization of nonzero dynamo coupling coefficients>, assume $c_1c_3\ne0$. Make the variable changes
$$
B=\widetilde B,\quad A=q e^{i\chi_A}\widetilde A,\quad V=v e^{i\chi_V}\widetilde V,
\qquad q=\frac1{\sqrt{|c_1c_3|}},\quad v=\sqrt{\frac{|c_3|}{|c_1|}},\quad\chi_V-\chi_A=-\arg c_1.
$$
The transformed coefficients are $\widetilde c_1=qv e^{i(\chi_V-\chi_A)}c_1=1$ and $\widetilde c_3=(q/v)e^{i(\chi_A-\chi_V)}c_3$, of modulus one. Thus
$$
\boxed{c_1=1,\qquad c_3=e^{i\beta},\qquad\beta=\arg(c_1^{\rm original}c_3^{\rm original}).}
$$
The remaining phase is genuine; the two complex coefficients cannot both generally be made positive real. A zero coupling cannot be normalized to one, so the printed normalization presupposes this generic nonzero-coupling case.
Now impose the specified simplification, in the normalized variables, and write $A=r e^{i\theta}$, $V=s e^{i\phi}$ with $r,s>0$ and signed real $B$. Put $\psi=\phi-\theta$. Separating real and imaginary parts gives
$$
\begin{aligned}
\dot B&=(\mu_1-es^2)B+2rs\cos\psi,\\
\dot r&=\mu_2r+c_2sB\cos\psi,\\
\dot s&=(\mu_3-d_4s^2)s+rB\cos(\beta-\psi),\\
\dot\theta&=\omega_2+\frac{c_2sB}{r}\sin\psi,\\
\dot\phi&=\omega_3+\frac{rB}{s}\sin(\beta-\psi),\\
\dot\psi&=\omega_3-\omega_2+\frac{rB}{s}\sin(\beta-\psi)-\frac{c_2sB}{r}\sin\psi.
\end{aligned}
$$
Constant moduli require the first three right sides to vanish. Since $rs\ne0$, the first equation also fixes $\cos\psi$, so a continuously evolving phase difference must be constant. Hence the last right side must vanish as well. Both phases then advance at one common frequency, giving a <relative equilibrium>. These four algebraic conditions and the displayed common-frequency equations govern all the requested nonzero constant-modulus solutions, without needing to solve them.
The pure-velocity branch is
$$
B=A=0,\qquad V=s_0e^{i(\omega_3t+\phi_0)},\qquad s_0^2=\frac{\mu_3}{d_4}>0.
$$
This presumes $d_4\ne0$; the usual saturated branch has $\mu_3,d_4>0$. Near it, set $A=e^{i(\omega_3t+\phi_0)}(u+iv)$, $B=b$, and define $\Delta=\omega_2-\omega_3$ and $\kappa=\mu_1-es_0^2$. Magnetic feedback in the velocity equation is quadratic, so the first magnetic <linearization> is
$$
\frac{d}{dt}\begin{pmatrix}b\\u\\v\end{pmatrix}
=\begin{pmatrix}\kappa&2s_0&0\\c_2s_0&\mu_2&-\Delta\\0&\Delta&\mu_2\end{pmatrix}
\begin{pmatrix}b\\u\\v\end{pmatrix}.
$$
A branch of the stated relative equilibria has a zero growth rate in this rotating frame. Its determinant therefore gives the <magnetic onset on a rotating velocity branch> condition
$$
\boxed{\left(\mu_1-e\frac{\mu_3}{d_4}\right)
[\mu_2^2+(\omega_2-\omega_3)^2]
=2c_2\mu_2\frac{\mu_3}{d_4}.}
$$
For $\mu_2^2+\Delta^2\ne0$, the zero-mode relation is $u+iv=-c_2s_0b/(\mu_2+i\Delta)$, so a nonzero $b$ gives a nonzero $A$ when $c_2\ne0$. With a transverse <eigenvalue> crossing and nondegenerate nonlinear saturation this is the generic magnetic bifurcation surface; magnetic reversal pairs its two signs. Degenerate cases require separate treatment. For example, if $\mu_2=\Delta=0$ and $c_2\ne0$, determinant zero is automatic but the stationary $u$ equation forces $b=0$, so that zero mode does not establish the requested branch with both magnetic amplitudes nonzero.
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