Why does exchanging coordinates produce a phase of $pm 1$ in an identical particle wavefunction?

Consider a system of two identical particles described by a wavefunction $psi(x_1, x_2)$. There are two kinds of exchange operators one can define:

• Physical exchange $P$, i.e. swap the positions of the particles by physically moving them around.
• The formal coordinate exchange $F$, where $F, psi(x_1, x_2) = psi(x_2, x_1)$.

Since $F^2 = 1$, the eigenvalues of $F$ are $pm 1$. Some books incorrectly say this proves that only bosons or fermions can exist. This is wrong because the argument also works in 2D, where anyons exist.

The real argument is to consider the eigenvalues of $P$, which are $pm 1$ only in three dimensions due to topology. In the 3D case, wavefunctions with $P$ eigenvalue $+1$ describe bosons, and those with $P$ eigenvalue $-1$ describe fermions.

However, all treatments of bosons and fermions say that bosons have $F$ eigenvalue $+1$, and fermions have $F$ eigenvalue $-1$. For example, you’ll see the equation
$$psi(x_1, x_2) = -psi(x_2, x_1)$$
for fermions. I’m not sure where this comes from; I accept the $P$ eigenvalue is $-1$, but as far as I can tell $F$ and $P$ are totally distinct. In particular, their eigenvalues must be different in two dimensions.

For identical particles in 3D, why are the $F$ and $P$ eigenvalues the same?