quantum mechanics - Can spin be related to a shift in angle?

If $\hat{T}(\Delta x) = e^{-\frac{i}{\hbar}\hat{p}\Delta x}$ is the spatial translation operator, then there exists a function $f$ from $\mathbb{R}$ to the ket space $V$ such that $\hat{T}(\Delta x) f(x) = f(x+\Delta x)$. Namely, the function that sends $x$ to the position eigenstate $|x\rangle$.

Similarly if $\hat{U}(\Delta t) = e^{-\frac{i}{\hbar}\hat{H}\Delta t}$ is the time evolution operator (for time-independent Hamiltonians), then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{U}(\Delta t) f(t) = f(t+\Delta t)$. Namely, the function that sends $t$ to $|ψ(t)\rangle$, the state of a particle at time $t$.

And similarly if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}{\hat{L}_z\Delta \theta}}$ is the orbital rotation operator, then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$. Namely the function which sends $\theta$ to $|r,\theta,\phi\rangle$, an eigenstate of the position operator $\hat{\theta}$ in spherical coordinates.

But my question is, if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}\hat{J}_z\Delta \theta}$ is the intrinsic rotation operator, then does there exist a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$? By intrinsic rotation operator I mean the rotation operation related to spin angular momentum.

I suspect the answer is no, because there is no operator corresponding to $\theta$, the parameter of the intrinsic rotation operator, since in quantum mechanics it doesn't really make sense to think of spin as a particle rotating about its own axis. But then again, there is no time operator in non-relativistic quantum mechanics, and yet the time evolution operator satisfies the property.

In any case, assuming that the answer to my question is no, I'd like a formal proof that there cannot exist such an $f$.

EDIT: I posted a follow-up question here.