When you consider a rotating frame of reference, you can assign a "rotational potential" to the centrifugal force.
If centrifugal force is
$$F_c = m\omega^2 r$$
then the centrifugal potential is
$$V_c = -\int F_c dr = -\frac12 m \omega^2 r^2\tag1$$
The potential energy as a function of deflection $\theta$ is given by
$$V_g = - m\;g\;\ell\;(1-\cos\theta)\tag2$$
For small angles $\theta$, we have that $r = \ell\sin\theta \approx \ell\theta$. This means that (1) can be written as
$$V_c=-\frac12 m \omega^2 \ell^2 \theta^2$$
Further, since $\cos\theta\approx1-\frac12\theta^2$ (2) can be written as
$$V_g = \frac12 m\;g\ell\;\theta^2$$
The sum of these is the effective potential well that the bob is moving in:
$$V_{eff} = \frac12m\ell(g-\omega^2\ell)\theta^2 $$ If this potential well has a minimum, the bob is stable; if it has a maximum, the bob is unstable. This follows from the sign of the expression in front of $\theta^2$, and thus we know the motion is stable if
$$g-\omega^2\ell \lt 0\\ \omega \gt \sqrt{\frac{g}{\ell}}$$
Note that in order to find the new steady state, you can't use the small angle approximation - you need to use the full $\cos\theta$ term to find the angle at which the conical pendulum will settle.
The assumption in the above is that the rotation of the rod will translate to the mass - in other words, that if there is a slight displacement of the mass it will rotate at $\omega$. Now if the bob is rotating while it is hanging vertically, that must mean that the rotation of the rod is being transmitted to the bob - which can only happen if the string does indeed transfer torque.