In the last chapter of John Preskill's Lecture Notes for Physics 219: Quantum Computation (PDF), he mentions the following on pg.12.
This behavior is compatible with the spin-statistics connection: the angular momentum $J$ of the $n$-anyon molecule satisfies
$$e^{-2\pi i J_n} = e^{-2\pi i n^2 J} = e^{i n^2 \theta}.$$
For example, consider a molecule of two anyons, and imagine rotating the molecule counterclockwise by $2\pi$. Not only does each anyon in the molecule rotate by $2\pi$; in addition one of the anyons revolves around the other. One revolution is equivalent to two successive exchanges, so that the phase generated by the revolution is $e^{i2\theta}$. The total effect of the two rotations and the revolution is the phase
$$\exp[i(\theta + \theta + 2\theta)] = e^{i4\theta}.$$
Not only does each anyon in the molecule rotate by $2\pi$; in addition one of the anyons revolves around the other. One revolution is equivalent to two successive exchanges so that the phase generated by the revolution is $e^{i2\theta}$.
How is this possible? A molecule here is just a localized region of two close anyons in the given space. Rotating the "molecule" should just be equal to revolving the two anyons one complete orbit about each other, that must give me a phase of $e^{i2\theta}$. What is meant by
Not only does each anyon in the molecule rotate by $2\pi$...
As far as I know, "rotation" of an anyon is no permutation in Topological QC. What's the concept involved here with additional $e^{i2\theta}$ phase that we are getting?