The answer to this question depends to some degree on your definition of 'motion'. There is a 'wave' around the nucleus, the values of which are complex numbers. For the standard basis of angular momentum eigenstates, the magnitude of these complex numbers is fixed over time, but the phases for many of the states rotate around the nucleus. Since global changes of phase make absolutely no difference to the physics (phase differences do have an effect, of course), it is questionable whether this constitutes anything physical 'moving', but if we treat the wavefunction phase as something 'real', then yes, you get waves spinning around the nucleus, just as Bohr thought.
The usual treatment separates out the time-dependence by solving the time-independent Schrodinger equation. The actual time-dependent wavefunction $\psi(t)$ is a complex exponenential of time multiplied by a fixed time-independent function $\psi_0$.
$$\psi(t,r,\theta,\phi) = e^{-iEt/\hbar}\psi_0(r,\theta,\phi)$$
And the $\psi_0$ has a complex exponential dependence on azimuth, $\phi$. For example, consider the $2p$-state:
$$\psi_{2,1,\pm 1}=\mp\frac{1}{8\sqrt{\pi}a_0^{3/2}}\frac{r}{a_0}e^{-r/2a_0}\sin\theta\, e^{\pm i\phi}$$
The final $e^{\pm i\phi}$ combines with the $e^{-iEt/\hbar}$ to yield a rotating movement of the phase in azimuth $e^{i(-Et/\hbar \pm \phi)}$, the rate of rotation being related to the energy of the orbital.
For higher order orbitals, the spherical harmonic part of the wavefunction is of the form:
$$Y^m_l=N\,P^m_l(\cos\theta)e^{-im\phi}$$
where $P^m_l$ is the associated Legendre functions which gives the dependence on polar angle (colatitude), and $e^{-im\phi}$ is the azimuthal dependence which, for order $m\neq 0$ rotates around the complex unit circle as discussed. (For $m=0$, the wavefunction sort of sloshes backwards and forwards along an axis; like a planet that falls directly towards a star and out the other side, then back again.)
This all works in exactly the same way as for pure basis states of linear momentum, for which the intuition may be better developed. A momentum eigenstate is a complex exponential for which the phase values move through space at a fixed velocity, called the phase velocity. However, the complex magnitude is constant, everywhere, and so the particle has no determined position. It is everywhere at once, equally, and continues to be so for all time. It is completely unlocalised, and so nothing appears to change as it moves.
We only get observable 'motion' when we combine momentum eigenstates to create a localised wavepacket, the waves of a range of wavelengths combining to yield a moving lump of probability. This lump moves at a different velocity to the phases, called the group velocity.
In just the same way, we can form angular momentum wavepackets that observably rotate around the nucleus by superposing pure states. For example, the posts here and here show how the superposition of two eigenstates yields an oscillating electron probability.
The pure angular momentum eigenstates behave very much like the pure linear momentum eigenstates, in that only the unobservable complex phase 'moves'. Complex magnitudes and hence probabilities are constant over all azimuths/positions, so motion makes no difference. And combining basis eigenstates to create a lump of probability results in motion at a completely different velocity to any of its constituents. Nevertheless, to whatever extent you consider that the complex phase is 'physically real', there is a real circulating motion around the nucleus associated with angular momentum eigenstates.
Those orbital diagrams are possibly a little misleading if they're not explained properly. I think they're generally doing something like plotting the real part of the complex wavefunction, so you can see the periodicity and structure. But the azimuthal dependence being of the form $e^{\pm im\phi}$ means that we are only seeing part of the full picture.