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All finite groups acting effectively on $S^2$ are conjugate to subgroups of $O_3$, the group of linear isometries of $S^2$. So your problem reduces to linear algebra, checking eigenvalues, and you have found representatives of all the conjugacy classes.

I'm not sure who is the first to prove the result I'm quoting but nowadays it follows immediately from 2-dimensional manifold+orbifold geometrization.

I think for involutions on $S^2$ the proof isn't so hard. If there's no fixed points the quotient is projective space and you're done. If there's fixed points use an equivariant tubularneighbourhood of the fixed point set and you've decomposed your manifold into either a circle + two discs (your reflection action) or two discs and an annulus (your rotation action). Either way, you're done. So the main ingredients in the argument are knowing 1) fixed points sets of finite group actions on manifolds are manifolds and have equivariant tubular neighbourhoods. 2) the classification of 2-manifolds. There are some combinatorial arguments I'm skipping like how you can rule out more than one circle as fixed point set, or anything other than two points, etc.

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All finite groups acting effectively on $S^2$ are conjugate to subgroups of $O_3$, the group of linear isometries of $S^2$. So your problem reduces to linear algebra, checking eigenvalues, and you have found representatives of all the conjugacy classes.

I'm not sure who is the first to prove the result I'm quoting but nowadays it follows immediately from 2-dimensional manifold+orbifold geometrization.

I think for involutions on $S^2$ the proof isn't so hard. If there's no fixed points the quotient is projective space and you're done. If there's fixed points use an equivariant tubularneighbourhood of the fixed point set and you've decomposed your manifold into either a circle + two discs (your reflection action) or two discs and an annulus (your rotation action). Either way, you're done.

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All finite groups acting effectively on $S^2$ are conjugate to subgroups of $O_3$, the group of linear isometries of $S^2$. So your problem reduces to linear algebra, checking eigenvalues, and you have found representatives of all the conjugacy classes.

I'm not sure who is the first to prove the result I'm quoting but nowadays it follows immediately from 2-dimensional manifold+orbifold geometrization.

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