Let $(M,g)$ denote a compact smooth Riemannian manifold with boundary and let $\mathscr F$ denote a family of smooth curves $\gamma$ such that they solve $$ \nabla^g_{\dot \gamma} \dot\gamma = F(\gamma,\dot{\gamma})$$ subject to $(\gamma(0),\dot{\gamma}(0)) \in TM$. Here $F$ is a smooth vector field on $M$. (As an example note that when $F\equiv 0$, then $\mathscr F$ consists of geodesics in $M$.)
Now, suppose we define conjugate points along curves in $\mathscr F$ in the natural way: For each $t$, $\xi \in TM\setminus 0$, we define $\exp_x (t,\xi)=\gamma(t)$ subject to $\gamma(0)=x$ and $\dot{\gamma}(0)=\xi$. Now we say that $\gamma(0)$ and $\gamma(t_1)$ are conjugate points if $D_{t,\xi}\exp_{x}(t,\xi)$ does not have maximal rank at $(t_1,\dot{\gamma}(0))$.
My question is as follows: Suppose that we assume that between any two points in $M$, there exists a unique curve in $\mathscr F$ that connects them. Does this imply that there are no conjugate points in $M$?
If the answer is no, can one add some mild conditions in addition to remove existence of conjugate points?
Thanks,