Problem. I'm working on an exercise from a text in Riemannian geometry, that tells me to do this:
Let $\mathbf{O}(n)=\{A\in\mathbb{R}^{n\times n}:A^t A=I\}$ be equipped with its usual (left-invariant!) Riemannian metric $g_p(X_p,Y_p)=\operatorname{tr}(X_p^t Y_p)$ for $p\in \mathbf{O}(n)$ and $X_p,Y_p\in T_p\mathbf{O}(n)$.
Show that for a $C^2$-curve $\gamma\colon I\to \mathbf{O}(n)$, it holds that $\gamma$ is geodesic if and only if $\gamma^t \ddot{\gamma}=\ddot{\gamma}^t \gamma$.
I think I solved it while trying to explain where I was stuck (it's always surprising how helpful it can be to just try to formulate a question), so I will post my suggested solution as an answer to this post, in case somebody else needs it or somebody else has something to add.