51st Ukrainian National Mathematical Olympiad, 4th Round

We will search for the function $f$ in the form $f(x)=kx$ with $k\ne 0$. Then $$\underset{n}{\underbrace{f(f(f\dots f(x)\dots ))}}=k^{n}x.$$ So the equality from the problem statement becomes: $$a_{1}kx+a_2{k}^{2}x+\cdots +a_{2011}{k}^{2011}x=x.$$ We cancel $x$ and obtain the following equation $$a_{2011}{k}^{2011}+a_{2010}{k}^{2010}+\cdots +a_2{k}^2+a_{1}k-1=0,$$ 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.