If we consider a geodesic $c: [a,b]\to M$ as a map $$f:\begin{cases} M\to M \\ c(a)\mapsto c(b)\end{cases}$$ is it true that the so defined map is an isometry?
To give some context:
A Riemannian manifold is said to be homogenous, if for any two points $p,q\in M$ there is an isometry $f\in \operatorname{Iso}(M)$ such that $f(p)=q$.
Propositon: If a Riemannian manifold is homogenous, it is geodesically complete.
Proof: If we take a normal neighbourhood (i.e. one for which there is an $\epsilon>0$ such that exponential map $\exp_p:B(0,\epsilon)\to M$ is diffeomorphic onto its image), since we can connect any two points by isometries, it follows that the same $\epsilon$ holds for all $p\in M$. In particular this enables us to extend the intervall of definition of any geodesic to all of $\mathbb{R}$ and thus $M$ is geodesically complete.
I was now wondering if the converse is also true, meaning that if we have a geodesically complete manifold, in particular we can connect any two points on a manifold by a geodesic, which if geodesics where isometries would imply that a complete manifold is homogenous.
Fixing a point the (time 1) exponential map is a map from the tangent space at that point to the manifold obtained by following a geodesic with a given start velocity. In general it is an isometry only along the geodesic.
As the tangent space is flat (zero curvature) in general the exponential map can not be isometric (if M has non-zero curvature). But there are other interesting properties. You may look in: Symmetric space
I don't really have any intuition for how it goes wrong. But curvatures are (magically) preserved under isometries so in general you can not hope for an isometry. (you may google for 'isometry and curvature')