Let $G$ be a non-connected compact Lie group and $Ad_g: G \rightarrow G, x \rightarrow gxg^{-1}$ be the adjoint action. Look at the induced homomorphism $ (Ad_g)_* : \pi_1 (G) \rightarrow \pi_1(G)$. If $g$ lies in the indentity component of $G$ , then $ (Ad_g)_*$ is the identity map. What kind of special property does $ (Ad_g)_*$ have if $g$ is not in the identity component of $G$ besides it is an isomorphism? Is it an nilpotent group action or even also the identity map? I bet this is not true in general but do not know a counterexample.
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$\begingroup$ Consider the semi-direct product $G$ of a product $H$ of circle groups with any finite subgroup of the automorphism group of $H$. $\endgroup$– Wilberd van der KallenCommented Apr 27, 2017 at 6:24
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$\begingroup$ I see, in your mentioned example, take $g=(id, f), f \in Aut (H)$ such that $f_*: \pi_1(H) \rightarrow \pi_1(H)$ is not identity (or not a nilpotent group action), then $(Ad_g)_*: \pi_1(G) \rightarrow \pi_1(G) $ is also not identity (or not a nilpotent group action), right? $\endgroup$– Xiaoyang ChenCommented Apr 27, 2017 at 7:59
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