Let $M$ be a smooth, connected manifold, and let $\nabla$ and $\nabla'$ be two covariant derivate operators on $M$ such that $\nabla \neq \nabla'$. As we know, a derivative operator $\nabla$ is defined by its set of geodesics, $\Gamma_\nabla = \{ \gamma\;|\;\xi^n\nabla_n\xi^a = 0 \}$, where $\xi$ is the tangent vector field of $\gamma$.
Under what conditions does there exist a diffeomorphism from $M$ to itself that maps geodesics to geodesics? Put differently, when does there exist a $d: M \to M$ such that $d^*\Gamma_\nabla = \Gamma_{\nabla'}$, where $d^*\Gamma_\nabla = \{ \gamma\;|\;d^{-1} \circ \gamma \in \Gamma_\nabla \}$.
When there exists such a diffeomorphism, is there a way to construct it explicitly? Note that for any two derivative operators $\nabla$ and $\nabla'$, there exists a smooth symmetric tensor field $C^a_{bc}$ such that
\begin{equation} (\nabla'_m - \nabla_m)\alpha^{a_1...a_r}_{b1...b_s} = \alpha^{a_1...a_r}_{n...b_s} C^n_{mb_1} + ... + \alpha^{a_1...a_r}_{b_1...n} C^n_{mb_s} - \alpha^{n...a_r}_{b_1...b_s}C^{a_1}_{mn} - ... - \alpha^{a_1...n}_{b_1...b_s}C^{a_r}_{mn} \end{equation}
So more specifically, supposing that $C^a_{bc}$ is known can we express the sought after diffeomorphism in terms of it?