I'm working on the proof for:
Let $(M,g)$ be a Riemannian manifold, $F:M \rightarrow M$ isometry, $N:=\{x \in M \vert F(x)=x\}$. Now I want to show that $N$ is a totally geodesic submanifold.
I know that this question has already been asked and I also tried to understand the following proof for this: 
but as far as I see it, this only proofs that $N$ is a submanifold and not that it is totally geodesic. Is this part of the proof somehow trivial? (If yes, can somebody maybe explain it to me..?)
Best Answer
I'll drop the * compared with your image.
A (unit-speed) geodesic in $M$ is uniquely specified by its starting point and a starting direction vector. We start a geodesic at $p\in N$ and a direction $v\in H$. The isometry fixes $(p,v)$ so fixes the geodesic. So $\exp_p(B(0,r)\cap H)$ is fixed.
The only thing left to prove is that $H\supseteq T_pN$. If $\gamma$ is a smooth curve on $N$ with $\gamma(0)=p$, then $\gamma$ is fixed, so $\gamma'(0)$ is invariant, so $T_pN\subseteq H$.