= Solution
The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,
$$
S(z+1)=f(z+1)f(-z)=-S(z).
$$
Its <modulus> is therefore periodic with period one. The assumed bound for $f$ on $0\leq\operatorname{Re}z\leq1$, 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 $f\equiv0$, a contradiction to that choice. Therefore $f\equiv0$ and $F=\Gamma$.
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