We will search for the function $f$ in the form $f(x) = kx$ with $k \neq 0$. Then
$$\underbrace{f(f(f\dots f(x)\dots))}_{n} = k^n x.$$
So the equality from the problem statement becomes:
$$a_1 kx + a_2 k^2 x + \dots + a_{2011} k^{2011} x = x.$$
We cancel $x$ and obtain the following equation
$$a_{2011}k^{2011} + a_{2010}k^{2010} + \dots + a_2k^2 + a_1k - 1 = 0,$$
which has a non-zero solution, because the left-hand side is a polynomial of an odd degree with non-zero leading and free coefficients. So, for this $k$ the function $f(x) = kx$ will solve the problem.