The idea is that if $\deg{P} \geq 2$, then $P(x)-x | P(P(x))-P(x)$, so $P(x)-x|P(P(x))-x$, thus a contradiction. Therefore the solutions are $\boxed{f(x) \equiv b}$ and $\boxed{f(x) \equiv ax+b}$ for some constant $a$ and $b$, which are possible, as desired.