Let M be a Riemannian manifold, $\gamma: I \rightarrow M$ be a curve which is parameterized with constant speed and $f:M\rightarrow M$ be an isometry with $f \circ \gamma = \gamma$. Furthermore $\ker(Id - d\_{\gamma(t)}f) = \mathbb{R} \dot{\gamma}(t), \forall t$. Show that $\gamma$ is a geodesic then.

I need some help with this. I don't know how f plays into this. We know that $d\_{\gamma(t)}f(a\dot{\gamma}(t)) = a\dot{\gamma}(t)$ for any $a \in \mathbb{R}$ , then for $ \mathbb{R} \ni c = \langle \dot{\gamma(t)}, \dot{\gamma(t)} \rangle = \langle d\_{\gamma(t)}f(\dot{\gamma}(t)), d\_{\gamma(t)}f(\dot{\gamma}(t)) \rangle$.

I don't see how any of this is supposed to be related to $\nabla\_{\dot{\gamma}}\dot{\gamma}$.

Why is it important that f is an isometry?