There is no such result even if we require that $F$ be fibre preserving.

If we require that $F:T^*M \to T^*M$ be fibre-preserving, then we have an induced diffeomorphism $f:M\to M$ on the base. Supposing that $F$ preserves the canonical 1-form $\theta$ on $T^*M$, we want to know whether $f^*=F$, i.e. whether $f^* \circ F^{-1}=id$. This can be verified locally. Hence it's enough to consider a linear automorphism $g: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ such that $g^*(\sum y_i dx_i)=\sum y_i dx_i$. The question is then: ``is $g$ necessarily the identity?" The answer is, very certainly, not.