For $2$ the answer is no. In fact, more generally, on any closed manifold there are always at least two points with two geodesics between them.

The reason is that every closed Riemannian manifold has at least one closed geodesic. I thought this result was due to Birhoff, but according to https://www.encyclopediaofmath.org/index.php/Closed_geodesic it's due to Lyusternik and Fet. (In the non-simply connected case, it's not too hard to prove and is due to Cartan).

Now, let $\gamma:[0,L]$ be a unit speed closed geodesic. We may assume wlog that $\gamma$ is minimal in the sense that if $\gamma$ is restricted to any subinterval of $[0,L]$, the resulting geodesic is *not* a closed geodesic.

Now, consider the points $\gamma(0)$ and $\gamma(L/2)$. If these are *not* the same point, then the geodesic $\gamma$ and "follow $\gamma$ backwards from $\gamma(0)$" are two geodesics between $2$ different points.

If, on the other hand, $\gamma(0) = \gamma(L/2)$ (but $\gamma'(0)\neq \gamma'(L/2)$, since otherwise we'd contradict minimality of $\gamma$), then we repeat the argument with $\gamma(0)$ and $\gamma(L/4)$. Eventually the sequence $\gamma(L/2), \gamma(L/4),...,\gamma(L/2^k)$ gets within the injectivity radius at $\gamma(0)$, and then the argument stops.