= Solution
Let $\delta_0=\theta_0-\varphi_0$ be $0$ or $\pi$, and write the angular displacements as $\eta_1,\eta_2$. To quadratic order the restoring <potential energy> is $\kappa\cos\delta_0(\eta_1-\eta_2)^2/2$. The linearized <normal mode> equation is
$$
\left[\kappa\cos\delta_0\begin{pmatrix}1&-1\\-1&1\end{pmatrix}-\omega^2\begin{pmatrix}I_1&0\\0&I_2\end{pmatrix}\right]\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=0.
$$
The <determinant> gives $\omega^2[\omega^2I_1I_2-\kappa\cos\delta_0(I_1+I_2)]=0$. Thus the <normal modes of two spring-coupled coaxial rotors> are
$$
\boxed{\omega_0^2=0,\qquad(\eta_1,\eta_2)\propto(1,1),}
$$
$$
\boxed{\omega_{\mathrm{rel}}^2=\kappa\left(\frac1{I_1}+\frac1{I_2}\right)\cos\delta_0,\qquad(\eta_1,\eta_2)\propto(I_2,-I_1).}
$$
For the aligned family, the relative mode oscillates with positive frequency
$$
\Omega=\sqrt{kaA\left(\frac1{ma^2}+\frac1{MA^2}\right)}.
$$
For the antipodal family, $\omega_{\mathrm{rel}}^2=-\Omega^2$, so that displacement has solutions $e^{\pm\Omega t}$ and the equilibrium is unstable.
The common <zero mode> has $\eta_1=\eta_2=C+Dt$. It describes a neutral displacement or uniform common <rotation>, not a restoring oscillation. Thus the aligned family is stable with respect to relative motion and has <orbital stability> modulo common rotations. A particular static point is not strictly <Lyapunov stable> against unrestricted small velocity perturbations: arbitrarily small nonzero $J$ eventually moves the common angle far from that point. With $J=0$, the common angle is fixed and the aligned relative <equilibrium point> is stable. This distinguishes the symmetry-induced drift from the exponential instability of the antipodal family.
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