Isometry that translates a geodesic has constant displacement In a Riemannian manifold $M$ we are given an isometry $\alpha$ that translates a geodesic $\gamma$, meaning that $\alpha(\gamma) \subseteq \gamma$, I would like to show that $d(\gamma(t), \alpha\gamma(t))$ is constant for all $t$.
I tried letting $s > 0$ such that $\alpha\gamma(0) = \gamma(s)$ and trying to prove that $d(\gamma(t), \alpha\gamma(t)) = d(\gamma(0), \gamma(s)) = s$ (supposing unit length parametrization) but so far I'm quite stuck, any idea on how to proceed?
 A: This is in general not true. Think of $M = \mathbb R$ with the standard metric and $\alpha (x) = -x$. If $\gamma$ is the geodesic given by $\gamma(t) = t$, then
$$ d(\gamma (t), \alpha(\gamma(t))) = 2|t|$$
depends on $t$.
A: On the positive side:
Suppose that $M$ is a complete, simply-connected Riemannian manifold of sectional curvature $\le 0$ (a Hadamard manifold). Then for every complete geodesic $c$ in $M$ and every isometry $\alpha$ of $M$ preserving $c$, either $\alpha$ acts on $c$ as a translation (in particular, has constant displacement) or as a reflection.
Edit. I will assume that $c$ has unit speed. By the classification of isometries of the real line, the action of $\alpha$ on $c$ is either a map
$$
c(t)\mapsto c(t+T),
$$
for some fixed $T$ (depending on $\alpha$) or is a reflection which, up to reparameterization, is $t\mapsto -t$.
Now, by the Cartan-Hadamard theorem, all geodesics in Hadamard manifolds are distance-minimizers and, thus,
$$
d(c(t), \alpha c(t))= d(c(t), c(t+T))=T. 
$$
A: Even if the isometry $\alpha$ is orientation-preserving, this fails. Let $\gamma$ be the intersection of the 2-sphere with the $xy$-plane, and let $\alpha$ be a rotation of angle $\pi$ about the $x$-axis. Then $\alpha$ preserves $\gamma$ but $d(\gamma(t),\alpha\gamma(t))$ takes all values in $[0,\pi]$.
