 Solution

No: the round sphere supplies a counterexample. On the unit round $S^2\subset {\mathbb{R}}^3$, restrict the smooth one-formThe antipodal map $A(x)=-x$ satisfies $A^{\ast }\theta =-\theta$, since $x_3^2$ is unchanged and $d{x}_1$ changes sign. Every unit-speed closed geodesic is a great circle, with $\gamma (t+\pi )=A(\gamma (t))$. Therefore its two half-circle integrals cancel:The same holds for all repeated traversals and either orientation.

However, its exterior derivative iswhich is not zero on the sphere. For example, at $p=(0,1/\sqrt{2},1/\sqrt{2})$ the tangent vectors $u=(1,0,0)$ and $w=(0,1/\sqrt{2},-1/\sqrt{2})$ give $d{\theta }_p(u,w)=1$. Hence $\theta$ is not a closed differential form, and cannot be an exact differential form, because $d(dh)=0$. This is the counterexample to geodesic one-form injectivity on the round sphere; periodic geodesic integrals alone do not imply exactness without an additional geometric hypothesis.