It's easy (2)

Since $f$ is its own inverse, $f(f(x))=x$ . If the leading term of $f(x)$ is $ax^n$ , the leading term of $f(f(x))$ is $a^2 x^{n^2}$ . We immediately see two things: firstly, $f$ is linear; secondly, its leading coefficient is $\pm1$ .

Having established these properties of $f$ , it's a simple matter to check the only solution is $f(x)=11-x$ , so the answer is $\boxed{11}$ .