Yes. Take $f(x) = x^{2007} - x^{2006} + 1$. Then clearly for all $n$ we have $n,f(n)$ coprime. To show $n$ and $f^m(n)$ are coprime for any $n$, note that $f(n) \equiv 1 \pmod{n}$. Additionally, note that if $k \equiv 1 \pmod{n}$ then $f(k) \equiv 1 \pmod{n}$. Hence the result follows that this polynomial works, because $f^m(n) \equiv 1 \pmod{n}$.