Time evolution of a quantum state is defined by the Hamiltonian $\hat{H}$ through its role in the Schrödinger equation $i\,\hbar\,\mathrm{d}_t\,\psi =\hat{H} \,\psi$ in the Schrödinger picture, or the observable evolution equation $\mathrm{d}_t\,\hat{A} = \frac{i}{\hbar}[\hat{H},\,\hat{A}]$ in the Heisenberg picture. This is the definition of time evolution in quantum mechanics.
So, unless the Hamiltonian's action on your state $\psi$ is related to the action of $\hat{L}_z$ by $\hat{H}\,\psi = \omega\,\hat{L}_z\,\psi$, the uniform rotation $\exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\psi$ is not a valid time evolution and $\exp\left(-\frac{i\,\theta}{\hbar}\,\hat{L}_z\,\right)\,\psi$ simply defines a co-ordinate rotation.
One instance where things do indeed work as you are thinking is when we think of Maxwell's equations as the one-photon Schrödinger equation, as I discuss in my answer here and here. Here the one-photon quantum state is uniquely defined by a vector field (the pair of positive frequnecy parts of the Riemann-Silbertein vectors $\vec{F}^\pm = \sqrt{\epsilon}\vec{E}\pm i\sqrt{\mu}\vec{H}$) and the one-photon Schrödinger equation is then:
$$i\,\hbar\,\partial_t\,\vec{F}^\pm = \pm\,\hbar\,c\nabla\times\vec{F}^\pm$$
(Note that here $\vec{E}$ and $\vec{H}$ do not stand for electric and magnetic fields, simply vector fields that define the photon's state just as the Dirac equation defines the first quantised electron's state as a spinor field). In momentum co-ordinates (wavevector space), this equation for a plane wave with wavevector $(k_x,k_y,k_z)$becomes:
$$i\,\hbar\,\partial_t\,\vec{F}^\pm= \pm\,\,c\,\left(k_x\,\hat{L}_z+k_y\,\hat{L}_y+k_z\,\hat{L}_z\right)\,\vec{F}^\pm$$
where here the angular momentum operators are of course:
$$\hat{L}_x=i\,\hbar\,\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)\;\;\hat{L}_y=i\,\hbar\,\left(\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right)\;\;\hat{L}_z=i\,\hbar\,\left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}\right)\;\;$$
So that if the wavevector is directed along the $z$ axis, you do indeed have $\vec{F}^\pm = \exp\left(-\frac{i\,\omega\,t}{\hbar}\,\hat{L}_z\,\right)\,\vec{F}^\pm$ where $\omega = k\,c$, defining left and right hand circular polarisation states.
I think the trick here is to note that the operator in (b) measures the component of angular momentum along the axis $\hat{n} = (3/5, 4/5, 0)$.
It's eigenvalues must be $\{2,1,0,-1,-2\}$, the same as those of $L_z$, because you could have chosen your z-axis to lie along $\hat{n}$.
Similarly, the operator in (c) is 7 times the component of $\vec{L}$ along the normalized axis $\hat{n} = (2/7, -6/7, 3/7)$.
It's eigenvalues must therefore be $\{14,7,0,-7,-14\}$, by the same reasoning.
Best Answer
In QM, angular momentum is conserved as an operator.
Deal with it.
This is best seen in the Heisenberg picture, where the equation of motion for $\hat{\mathbf L}$ in a spherically-symmetric hamiltonian $\hat H$, $$ i\hbar\frac{\mathrm d}{\mathrm dt}\hat{\mathbf L} = [\hat{\mathbf L},\hat H] = 0, $$ so $\hat{\mathbf L}$ is conserved as an operator as $t$ goes from $t=0$ onwards. In particular, this means that:
These conclusions also hold equally well in the Schrödinger picture.
(On the other hand, if you know that the system is in some eigenstate $|\psi⟩$ of $\hat L^2$ and $\hat L_z$ and it evolves under a spherically symmetric hamiltonian $\hat H$ in the Schrödinger picture, you can conclude that it will remain as an eigenstate of $\hat L^2$ and $\hat L_z$ with its same eigenvalues, but you cannot conclude that the state does not change, as there is typically radial motion which can obviously change.)