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Hello,

I would like to know whether the only non trivial involutary group isomorphism $f:(\mathbb{C}^{*},\times)\to(\mathbb{C}^{*},\times)$ is the complex conjugation or not. Thank you in advance.

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closed as too localized by Qiaochu Yuan, Martin Brandenburg, quid, André Henriques, Andreas Thom Jul 22 '11 at 14:32

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See ccc's answer here for a complete list of all (measurable) endomorphisms. You can easily check which ones of those are involutive: math.stackexchange.com/questions/52319 –  Theo Buehler Jul 21 '11 at 21:21
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2 Answers 2

$z\mapsto \frac{1}{z}$

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Indeed...Maybe it's time for me to go to bed :/ –  Sylvain JULIEN Jul 21 '11 at 20:41
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How about $$ z\mapsto \frac{z}{|z|^2}?$$

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