= Solution
Again, no. On the unit round sphere and for a sufficiently small nonzero $\varepsilon$, consider
$$
\gamma(\theta)=
\frac{(\cos\theta,\sin\theta,\varepsilon\sin3\theta)}
{\sqrt{1+\varepsilon^2\sin^23\theta}},
\qquad0\leq\theta\leq2\pi.
$$
This is a smooth embedded closed curve and may be reparametrized by arc length. It obeys
$$
\gamma(\theta+\pi)=-\gamma(\theta),
$$
so the antipodal <isometry> preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry $S_1\to S_2$.
The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with $\cos\theta$ and $\sin\theta$. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles, $\gamma$ is not geodesic. This <antipodally symmetric nongeodesic spherical curve> shows that the point-transitivity assumption cannot be omitted.
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