Ukrainian Mathematical Olympiad

Let us find the explicit form of the function $y=f(x)$. Let $P(p;p^2)$ be an arbitrary point of the parabola $y=x^2$, and $Q(q;f(q))$ be the point on the graph $y=f(x)$, symmetric to $P$ with respect to the point $(1;1)$. Then: $$\frac{p+q}{2}=1,\frac{p^2+f(q)}{2}=1.$$ Eliminating $p$ from these equations, we get $f(q)=-q^2+4q-2$.

Next, we need to solve the equation $f(f(x))=f(x)$. We have: $$-(f(x){)}^2+4f(x)-2=f(x),(f(x){)}^2-3f(x)+2=0,(f(x)-1)(f(x)-2)=0,$$ $$\left[\begin{array}{cc}f(x)=1,& -x^2+4x-2=1,\\ f(x)=2,& -x^2+4x-2=2.\end{array}\right]$$ It remains to solve the obtained quadratic equations.

Answer: $1$, $2$, $3$.