I would know if it is possible to create a natural parallel transport on a Riemannian manifold $M$ with the following procedure.


Define a minimum length curve (m.l.c.) $\gamma$ from $P$ to $Q$ requiring to minimize the functional of length $L(\gamma)$ and show that for every two points $P$ and $Q$ in $M$ exists such m.l.c. and is unique .


**Parallel transport**


Take two points $P$ and $Q$ in $M$ and a vector $v$ in $T\_P M$ consider a m.l.c. $\gamma$ 
from $P$ to $Q$ and prove that there is a only vector field $V$ on $\gamma$ such that $V(P)=v$ and the angle between $V$ and $\gamma '$ is constant. Now we set $w=V(Q)$ as the parallel transport of $v$ in point $Q$.


I would know if the required proofs can be done (eventually in under some conditions) or if this procedure is in some textbook. I think that this porcedure is a natural way to construct a parallel trasport and maybe this parallel transport gives us also a Levi-Civita connection.



Parallel transport with minimum length curve
Your assumption that minimal geodesics exist and are unique, in the complete case, is equivalent to $M$ being simply connected and having no conjugate points. This is already quite a restrictive class of manifolds.


In your above comment, you suggest that, in the case where multiple minimal geodesics between points exist, perhaps parallel transport is independent of the path taken. But that's typically not the case. For example, consider two antipodal points $p$ and $\tilde{p}$ on a two-dimensional sphere $S^2$. If you connect them via arcs of great circles $\gamma\_1$ and $\gamma\_2$ with initial vectors $v\_1$ and $v\_2$, respectively, that are perpendicular at $p$, then it's easy to see that parallel transporting $v\_1$ around $\gamma\_1$ and $\gamma\_2$ produces vectors at $\tilde{p}$ that point in opposite directions.


Finally, while the angle between $V$ and $\gamma'$ is constant under parallel transport, that's not nearly restrictive enough to characterize it in dimension three or higher, because one can rotate around $\gamma'$.


