jmc

Since $f(2)=f^{-1}(2)$, we can substitute $f^{-1}(2)$ freely for $f(2)$. Therefore, $f(f(2))=f(f^{-1}(2))$, which is $\boxed{2}$ (since $f(f^{-1}(x))=x$ by definition).

Notice that we didn't actually need the value $4$ given in the problem.