I will work in ${\rm GL}$ instead of ${\rm PGL}$.

The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of rank 3 over $\mathbb{Z}$ there are up to isomorphism, where $G$ is the cyclic group of order 3. Any such module is a direct sum of indecomposable modules, and those have been classified in I. Reiner, Integral representations of cyclic groups of prime order, Proc. Amer. Math. Soc. 8 (1957), 142–146. There are three indecomposable $\mathbb{Z}[G]$-modules:

- the trivial module of $\mathbb{Z}$-rank 1, $\Gamma_1$,
- the augmentation ideal of $\mathbb{Z}[G]$, which has $\mathbb{Z}$-rank 2, $\Gamma_2$,
- the regular module $\Gamma_3=\mathbb{Z}[G]$ itself.

So there are two isomorphism classes of faithful modules of rank 3:

- $\Gamma_1\oplus \Gamma_2$,
- $\Gamma_3$.

One should beware, that, in general, the Krull-Schmidt Theorem fails for $\mathbb{Z}[G]$-modules, but in this case it is easy to see that the two guys are not isomorphic, since e.g. in one of them the trivial isotypical component is a direct summand, and in the other one it isn't. Alternatively, the two are not isomorphic over $\mathbb{Z}_3$, and Krull-Schmidt *does* hold over local rings.

${}^1$ To remove the word "essentially", one needs to check that any matrix obtained in this way is conjugate to its inverse. This is true because each of the indecomposable modules listed above can be extended to a module under the symmetric group $S_3$, as is easy to check.