Let $\mathbb{S}^n$ be the $n$-dimensional unit sphere, equipped with the standard round Riemannian metric.
Let $f:\mathbb{S}^n \to \mathbb{S}^n$ be a diffeomorphism and suppose that for every (parametrized) geodesic $\gamma$, $f \circ \gamma$ is also a (parametrized) geodesic.
Must $f$ be an isometry? (that is the restriction of an orthogonal matrix on $\text{O}(n+1)$).
An equivalent condition on $f$ is that $\nabla df=0$ where $\nabla=\nabla^{(T\mathbb{S}^n)^*} $ $ \otimes \nabla^{f^*T\mathbb{S}^n}$ is the relevant tensor product connection.
Best Answer
Yes, such an $f$ must be an isometry.
Let $p\in S^n$, $v\in T_p S^n$ with $v\neq 0$. By polarization, it is enough to show that that $\|d_p f(v)\|= \|v\|$.
So, let's establish this equality. To do so, first note that $\gamma:\mathbb{R}\rightarrow S^n$ defined by $\gamma(t) = \exp_p(tv)$ is injective on $\left[0,\frac{2\pi}{\|v\|}\right)$.
Because $f$ is a diffeomorphism, $f\circ\gamma$ is injective on the same interval. But since $f$ maps geodesics to geodesics, we know $f\circ \gamma = \exp_{f(p)}(td_p f(v))$. This implies that $\|d_p f(v)\| \leq \|v\|.$
Further, because $\gamma$ fails to be injective on $\left[0, \frac{2\pi}{\|v\|}\right]$, the same must be true of $f\circ \gamma$. This implies that $\|d_pf(v)\| \geq \|v\|$. The two inequalities together now give the result.