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I think I misunderstand something about the definition of a homeomorphism of a topological space. We know an automorphism of a topological space $X$ is a homeomorphism from $X$ to itself. So it should be bijective and continuous with a continuous inverse. Now, my understanding is that every rotation of the circle is a homeomorphism of the circle. But in this question, the OP says that the automorphism group of $S^1$ as a topological group is $\mathbb{Z}_2$. What is wrong with my conclusion? Do rotations leave the group stucture of the circle unchanged?

Math learner
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    The key is "topological group." If you think of the circle as a group, then rotations don't preserve the identity, so are not group automorphisms. – Cheerful Parsnip Nov 16 '22 at 05:03
  • @CheerfulParsnip Thanks! But if we want to forget the group structure, rotations are homeomorphisms of the circle. So they are also automorphisms of $S^1$ as a topological space. Am I wrong? – Math learner Nov 16 '22 at 05:23
  • Yes. Rotations are topological automorphisms. – Cheerful Parsnip Nov 16 '22 at 05:27

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