Spin-$n$ particle comes back to itself after $360/n$ degree rotation On the Wikipedia page for spin, a claim is made that a spin-$n$ particle comes back to itself after a $360/n$ degree rotation.
I quote:

A spin-zero particle can only have a single quantum state, even after torque is applied. Rotating a spin-2 particle 180 degrees can bring it back to the same quantum state and a spin-4 particle should be rotated 90 degrees to bring it back to the same quantum state. The spin-2 particle can be analogous to a straight stick that looks the same even after it is rotated 180 degrees and a spin-0 particle can be imagined as sphere, which looks the same after whatever angle it is turned through.

My question is: Is this claim true? If it is false, what is the correct claim?

I believe the claim is false for $n>1$. I believe that for half-integer spin particles, no less than a $720$ degree rotation is required, and for integer spin no less than a $360$ degree rotation is required to bring the internal state back to itself. To explain my reasoning, consider the generators of rotation about the z-axis:
$S_z^{1/2} = \begin{pmatrix}
1/2 & 0\\
0 & -1/2
\end{pmatrix}$
$S_z^{1} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1 
\end{pmatrix}$
$S_z^{3/2} = \begin{pmatrix}
3/2 & 0 & 0 & 0\\
0 & 1/2 & 0 & 0\\
0 & 0 & -1/2 & 0\\
0 & 0 & 0 & -3/2
\end{pmatrix}$
$S_z^{2} = \begin{pmatrix}
2 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & -2 
\end{pmatrix}$
A rotation about the $z$-axis with angle $\theta$ is given by $e^{iS_z \theta}$.
Since the rotation matrices are diagonal, we have
$e^{iS_z^{1/2}\theta} = \begin{pmatrix}
e^{i\theta/2} & 0\\
0 & e^{-i\theta/2}
\end{pmatrix}$
$e^{iS_z^{1}\theta} = \begin{pmatrix}
e^{i\theta} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & e^{-i\theta} 
\end{pmatrix}$
$e^{iS_z^{3/2}\theta} = \begin{pmatrix}
e^{i3\theta/2} & 0 & 0 & 0\\
0 & e^{i\theta/2}& 0 & 0\\
0 & 0 & e^{-i\theta/2} & 0\\
0 & 0 & 0 & e^{-i3\theta/2}
\end{pmatrix}$
$e^{iS_z^{2}\theta} = \begin{pmatrix}
e^{i2\theta} & 0 & 0 & 0 & 0\\
0 & e^{i\theta} & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & e^{-i\theta} & 0 \\
0 & 0 & 0 & 0 & e^{-i2\theta}
\end{pmatrix}$
Clearly, we need $\theta = 4\pi$ for spin-$1/2$ and $\theta = 2\pi$ for spin-$1$ as the minimum nonzero rotation angle in order to get back to the identity matrix.
However, because $1/2$ is an element on the diagonal of $S^{3/2}_z$, we also need $\theta = 4\pi$ as the minimum nonzero rotation angle to get back to the identity matrix, not $\theta = 4\pi/3$. For spin-$2$, we need $\theta = 2\pi$ and not $\theta = \pi$ as minimum nonzero rotation angle in order to get back to the identity matrix.
Quite generally, the above seems to say that half-integer spins require a minimum rotation angle to "get back to themselves" of $\theta = 4\pi$, while (nonzero) integer spins require $\theta = 2\pi$.
The claim that the rotation angle is $\theta= 2\pi/n$ for spin-$n$ particles appears to be incorrect.
Though I feel mostly convinced by my arguments above, I have heard the Wikipedia page's claim more than once before, so I fear I am missing something, hence the motivation for my question.
 A: I did not originally answer this question because I felt someone else might be able to offer a more authoritative answer on the matter. But in the intervening time I have become more confident in the answer I am about to give, though if anyone sees any issues with it, please feel free to let me know. Also, sorry if this is a little long, but I wanted to try and give as complete an answer as I could to this question. The TL;DR is I'm pretty sure OP is correct, with a possible exception for massless particles under some conditions.
Let me begin by pointing out that there are actually two distinct notions of spin which are not quite the same. In both cases, the entire purpose of spin is to specify the representation of the little group of the Lorentz group. This is the standard definition as it appears from the Wigner classification of Lorentz group representations. You can find information about the Wigner classifications in a number of places, but Weinberg's QFT volume 1 is always a good option for this.
The little group itself comes in two varieties, corresponding to whether the particle in question is massive or massless. In the massive case, the little group is SO(3) (or its universal cover SU(2)) while in the massless case, the little group is ISO(2) (the group of Euclidean transformations in the plane). Those both SU(2) and ISO(2) representations can be labeled by something we can call "spin," they aren't quite the same and the difference will matter for the sake of this answer. Note that as a point of language, these two notions of spin are usually not verbally distinguished and are both referred to as "spin." Well, more properly the massless version should be called helicity, but I often hear people gloss this verbal distinction.
In any case, let me talk about the massive case first. As we well know, when looking at representations of SU(2), we first diagonalize $J_z$, whose eigenvalues are the (half-)integers between $\pm j$. The representations of SU(2) are then labeled by this largest eigenvalue $j$, and it's this $j$ that we refer to as the "spin" of the representation. From this point forward, OP's argument follows, and can indeed be generalized to the statement that for $j$ odd the periodicity will be $4\pi$ and for any $j$ even, the periodicity will be $2\pi$.
This can, in fact, be understood from a purely group theoretic standpoint as well (though direct calculation is clearly sufficient). As Lie groups, SU(2) is well-known to be the double-cover of SO(3). The manifold SO(3) itself is the three-sphere, and hence the coordinates on SO(3) are 2\pi periodic...this is nothing more than the statement that the Euler angles are...angles and hence are $2\pi$ periodic.
The SU(2) is a little more complicated, but the precise statement is that SU(2) winds twice around the SO(3) in the covering map...it's SU(2) which has $4\pi$ periodicities as a manifold and the covering map of SO(3) is essentially a $\mathbb{Z}_2$ quotient of SU(2). This is also the statement that half-integer spins are spinnor representations while integer spins are vector representations. As a group, SO(3) only has integer spin representations. The half-integer stuff only comes from the SU(2) cover.
Anyway, the long and short of it is that OP is correct about the periodicities when it comes to representations of SO(3) or SU(2). But as I've been setting up, ISO(2) representations are a little bit different.
So let's talk about ISO(2), the representations of which Michael Seifert alluded to in the comments. The Euclidean transformations in the plane essentially consist of translations and rotations in the plane. The subgroup of rotations in the plane are isomorphic to U(1), and it turns out that the representations of the U(1) subgroup label the representations of ISO(2)...which is also why it vaguely makes sense to still speak of spin. There's still a rotation of some kind going on. Again, the notion of "spin" here is properly referred to as helicity, but as an abuse of language it's often still called spin.
Now, the representations of U(1) may in general be labeled by any real number and there's no actual restriction on what the helicity may be in the same way there's no restriction on what value an electric charge may take (Dirac quantization not withstanding). The restriction on the possible values of the helicity comes only from global considerations on how ISO(2) as the little group embeds in the Lorentz group (or its cover, $SL(2,\mathbb{C})$). Those global considerations restrict the helicity $\sigma$ to to either integer or half-integer values.
But this is actually a little strange as it tells us a massless particle is labeled by a single number, $\sigma$, its helicity. Unlike in SO(3), we are not guaranteed to also get a particle of helicity $-\sigma$ in the spectrum of particles. Rather, this latter condition that massless particles should come in pairs with helicity $\pm\sigma$ only follows from CPT invariance of relativistic, unitary QFTs since the operation of CPT on helicity is to negate it. Without CPT, there would be no reason for a helicity $\sigma$ particle to even have a helicity $-\sigma$ counterpart!
The take-away of all this about massless particles is that if we only have the (half-)integers $\pm\sigma$, then the transformations
$$
e^{\pm i\sigma\theta}
$$
which act on massless particles are generically periodic modulo $2\pi/\sigma$. So this is the only case in which you can find rotations applied to a particle which wind the circle faster than $2\pi$. However, it's worth pointing out that so long as we are working in a domain where the Weinberg-Witten theorem is applicable, there is no way to have a massless particle of helicity greater than 1 anyway...so in all such cases massless particles also experience either $2\pi$ or $4\pi$ periodicity.
A: Invariance of all states of a half-integer spin requires an angle of $4\pi$. Invariance of the state with $S_z=m_z$ requires a rotation about the quantisation axis by only $2\pi/m_z$.
