A positive helicity photon with $\vec k = k\hat z$ has spin vector given by ($\hbar =1$):
$$ \vec s = - (\hat x + i\hat y) = \sqrt 2 Y_1^1(\theta, \phi)$$
It should be easy to show that is an eigenstate of rotations about the $z$-axis with eigenvalue (m=1):
$$ \lambda(\phi) = e^{im\phi} = e^{i\phi} $$
so that:
$$ \lambda(2\pi) = e^{2\pi i} = +1 $$
If I pick a transverse axis and (alias) rotate 90 degrees, I get:
$$ \vec s = -(\hat x' + \hat z') =
\sqrt 2 Y_1^1(\theta', \phi') - Y_1^0(\theta', \phi') - \sqrt 2 Y_1^{-1}(\theta', \phi') $$
A 180 degree rotation gives:
$$ \vec s = -\hat x'' + i\hat y'' = -\sqrt 2 Y_1^{-1}(\theta'', \phi'')$$
I chose an alias rotation to change the coordinates and leave the spin, momentum unchanged, so in the new coordinate system, that is still a +1 helicity photon, as the momentum becomes:
$$ R(90^{\circ})\hat z = -\hat y' $$
$$ R(180^{\circ})\hat z = -\hat z'' $$
What makes this odd is that I used cartesian vectors for the momentum, and implicit spherical vector for the spin. Spherical vectors are complex combinations of cartesian vectors that look exactly like spherical harmonics in cartesian coordinates. They are useful because the basis vectors are eigenvectors of z rotations with eigenvalues $\exp{im\phi}$ for $m \in (-1, 0, 1) $.
In all cases, a 360 degree rotation is returns the state to the the original state.
It's also odd, because in the natural basis:
$$ Y_1^0(\theta, \phi) = \hat z $$
is not available to the photon spin. $m=0$ is transverse polarization.
Note an $m=2$ spin-2 particle would have a spin dyad:
$$\vec s\vec s= [-\sqrt 3(\hat x + i\hat y)][-\sqrt 3(\hat x + i\hat y)]$$
$$ = 3\left[\begin{array}{ccc} 1 & i & 0 \\ i & -1 & 0 \\ 0 & 0 & 0 \end{array}\right] = 3Y_2^2(\theta, \phi) $$
You can verify that is an eigentensor of z-rotations with eigenvalue
$$ \lambda(\phi) = e^{2i\phi}$$
For fun, a 90 degree rotation would be:
$$ = 3\left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 0 & 0 \\ i & 0 & -1 \end{array}\right] $$
$$ \frac 3 2 Y_2^2 - 3iY_2^1-\sqrt{\frac{27} 2}Y_2^0 +3iY_2^{-1} +\frac 3 2 Y_2^{-2} $$
and a 180 degree rotation would return to the original state.
