 Solution

With $q=0$, expand $F(F(x))$ for $x=O(\sqrt{\mu })$ through $O({\mu }^{3/2})$. Besides zero, $F^2(x)=x$ gives $x^2=\mu /p+O({\mu }^2)$, and the two points are exchanged by $F$. The derivative product is $1-4\mu +O({\mu }^2)$, so the period-two orbit $|x|=\sqrt{\mu /p}+O({\mu }^{3/2})$ is stable.