 Solution

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,Its modulus is therefore periodic with period one. The assumed bound for $f$ on $0\le \mathrm{Re}z\le 1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb{C}$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an open set; the identity theorem would then give $f\equiv 0$, a contradiction to that choice. Therefore $f\equiv 0$ and $F=\Gamma$.