Suppose that we have a group $G = K\rtimes\mathbb{Z}$, where $G$ is finitely generated, but $K$ is not finitely generated, and let $\phi(1)$ be the automorphism of $K$ corresponds to $1_\mathbb{Z} \in \mathbb{Z}$, which defines the semi-direct product. Is it possible for $\phi(1)$ to have a fixed point that is not the identity in $K$?
I got this question when I was reading about the Lamplighter group, $L_2= \mathbb{Z}_2 \wr \mathbb{Z}$, in this case $\phi(1)$ "shifts" every element in $\bigoplus_{-\infty}^{\infty}\mathbb{Z}_2$ to the right by $1$. Then I was wondering if $G = K\rtimes\mathbb{Z}$ has the above properties, is it true that $\phi(1)$ can't have fixed points in general?
I was thinking that if we write out the generating set of $K=\langle k_1,\ldots,k_n \rangle $, where none of $k_i$ is the identity. It seems to me that $\phi(1)$ can't fix any of $k_i$.
Any help/counterexample would be really appreciated.
Let $T=\bigoplus_{n=-\infty}^{n=\infty}\mathbb{Z}_2$,
let $f:T\to T$ shift every element one step in the way you described,
and for the semi-direct product $G=(T\oplus\mathbb{Z})\rtimes\mathbb{Z}$,
let $\phi(1)(x,n)=(f(x),n)$.
Then $G$ is finitely generated, but $\phi(1)$ has a countably infinite number of fixed points.
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To see a different example, in which the fixed points of $\phi(1)$ are not even a finitely generated group, you can head over to https://mathoverflow.net/questions/418977/if-k-rtimes-mathbbz-is-a-finitely-generated-group-but-k-isnt-must-the-f.