= Solution
For an arbitrary smooth variation $q_i\mapsto q_i+\varepsilon\eta_i$ vanishing at both endpoints,
$$
\delta S=\int_{t_1}^{t_2}\sum_i\left(L_{q_i}\eta_i+L_{\dot q_i}\dot\eta_i\right)dt
=\int_{t_1}^{t_2}\sum_i\left(L_{q_i}-\frac d{dt}L_{\dot q_i}\right)\eta_i\,dt.
$$
The endpoint term in <integration by parts> is zero. <stationarity> for every variation and the <fundamental lemma of the calculus of variations> imply the <Euler-Lagrange equations> $d(L_{\dot q_i})/dt=L_{q_i}$. Least action here means stationary action, not necessarily a global minimum. The <conjugate momentum> is $p_i=L_{\dot q_i}$; a <cyclic coordinate>, $L_{q_i}=0$, makes $p_i$ constant.
For the <symmetric top>, $\psi$ and $\phi$ are cyclic. Thus
$$
p_\psi=I_3(\dot\psi+\dot\phi\cos\theta)=I_3\omega_3,\qquad
p_\phi=I_1\dot\phi\sin^2\theta+I_3\omega_3\cos\theta
$$
are constant, and consequently so is $\omega_3$. <Conservation of energy from time-translation invariance> also supplies the conserved <energy>
$$
E=\frac12I_1(\dot\theta^2+\dot\phi^2\sin^2\theta)+\frac12I_3\omega_3^2+V(\theta).
$$
These give the requested additional conserved quantities $p_\phi$ and $E$. The $\theta$ equation is
$$
I_1\ddot\theta=I_1\dot\phi^2\sin\theta\cos\theta-I_3\omega_3\dot\phi\sin\theta-V'(\theta).
$$
For constant $\theta$ and constant <precession> rate $\dot\phi$, it reduces to
$$
\boxed{I_1\dot\phi^2\sin\theta\cos\theta-I_3\omega_3\dot\phi\sin\theta=V'(\theta).}
$$
Conversely, choose constants satisfying this balance and set $\dot\psi=\omega_3-\dot\phi\cos\theta$. The cyclic equations and $\theta$ equation are then all satisfied, so this constructs uniform <precession> wherever the Euler-angle chart is regular.
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