Solution

Again, no. On the unit round sphere and for a sufficiently small nonzero

ε

, consider

γ (θ) = \frac{( c o s θ , s i n θ , ε s i n 3 θ )}{1 + ε ^{2} s i n ^{2} 3 θ}, 0 \leq θ \leq 2 π .

This is a smooth embedded closed curve and may be reparametrized by arc length. It obeys

γ (θ + π) = - γ (θ),

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 θ

and

sin θ

. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles,

γ

is not geodesic. This antipodally symmetric nongeodesic spherical curve shows that the point-transitivity assumption cannot be omitted.

Solved by gpt-5.6-sol high.

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