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The following question is inspired by this post, where it is proven that given $X$ a compact Hausdorff group, there is a one-to-one correspondence between automorphism of $C(X)$ and homeomorphisms defined from $X$ to $X$. My question is: when $X$ is a compact Hausdorff topological group, is there any correspondences between the automorphism group of $C(X)$ and homeomorphic automorphism of $X$?

Below is an example that may contribute to a counter-example to some correspondence that is not one-to-one. Consider the unit circle $\mathbb{T}$. In this post, it is proven that $\operatorname{Aut}(\mathbb{T})$ as a topologicall group is isomorphic to $\mathbb{Z}_2$, which means $\mathbb{T}$ only has two group autormophisms, the identity mapping and the conjugate mapping. I am tempted to believe the automorphisms of $C(\mathbb{T})$ has more than two elements. However, after I found out the mapping $f(z) \mapsto f(\overline{z})$ coincides with $f(z) \mapsto \overline{f(z)}$, I cannot find the third one.

Kaku Seiga
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    No, of course not; by the result you cite, the automorphism group of $C(X)$ is the homeomorphism group of $X$. So the automorphism group of $C(\mathbb{T})$ is the homeomorphism group of $\mathbb{T}$, e.g. you can rotate the circle. To have a chance of getting the automorphism group of $\mathbb{T}$ you need additional structure that corresponds to the group structure of $\mathbb{T}$, e.g. some version of convolution or some comultiplication-like operation. – Qiaochu Yuan Feb 11 '25 at 23:24
  • Thank you for your inputs. You reminded me that an automorphism of $C(\mathbb{T})$ can induce an mapping defined on the space of multiplicative linear functionals of $C(\mathbb{T})$, which is $\mathbb{T}$ again. Now $C(\mathbb{T})^*$ can be endowed with the Arens product but for now I cannot see the Arens product coincides with the usual product on $\mathbb{T}$. – Kaku Seiga Feb 12 '25 at 00:00

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