To show that there is a fixed point, select $a^2 - b^2 = p$ for large prime $p$ and naturals $a,b$ meaning $a= \tfrac{p+1}{2}$, $b=\tfrac{p-1}{2}$ so you have $f(a) + f(b) \mid f(a)(a^2 - b^2)$ and $\gcd(p=a^2 - b^2 , f(a) + f(b))$ cannot be one due to size reasons, since $f(a) + f(b) \le a + b = p$ so we have equality everywhere i.e. $f(a) + f(b) = a+b = p$, ergo $f(a) = a$ and $f(b) = b$. To finish fix any natural $x$ and choose large enough $a$, $f(x) + a \mid (a^2 - x^2)f(x) \implies f(x) + a \mid (f(x)^2 - x^2)f(x)$ meaning $f(x)^2 = x^2$ or $f(x)=x$.