If a finitely generated semi-direct product $\mathbb{Z}$ acting on a non finitely generated group, can the fixed points form a non-normal subgroup? Suppose that we have a finitely generated and residually finite group $G = K\rtimes\mathbb{Z}$, but $K$ is not finitely generated. Let $T$ be a finite subset of $K$ such that $\langle (0,1), (k,0)\mid k \in T \rangle$ is a generating set of $G$. Let $H =  \langle T  \rangle$, a finitely generated subgroup of $K$.
Now, suppose in addition, that the automorphism satisfies that $ H \not\subset \phi(1)(H) $ and  $ H \not\subset \phi(-1)(H)$. (A example of group satisfies this property is the wreath product $H \wr \mathbb{Z}$)
Question: Is it possible that the fixed points of $\phi(1)$, which is the automorphism on $K$ corresponding to $1_\mathbb{Z}$, does not form a normal subgroup of $K$.
My thought so far: Cleary $K$ cannot be abelian (hence we can't construct such a group using wreath product). Also, there is another group given by Derek (https://mathoverflow.net/a/419028/479955) in a related question, in this case, $K$ is not abelian, but $Z(K)$, the centre of $K$, consists exactly all the fixed points, and so the set of fixed points is a normal subgroup of $K$.
Any help/idea to construct such a group would be really appreciated.
 A: How about the group defined by the presentation
$$\left\langle x_i\,(i \in {\mathbb Z}),y,z \mid
x_i^3= [x_i.x_j]= (x_iy)^2 = 1, x_i^z = x_{i+1}\, (\forall i,j \in {\mathbb Z}),y^z=y \right\rangle$$
with $K = \langle x_i\,(i \in {\mathbb Z}),y \rangle$ and $T = \{ x_0,y\}$. So $\phi(1)$ fixes $y$ and maps $x_i$ to $x_{i+1}$ for all $i$.
(This contains $C_3 \wr {\mathbb Z}$ as a subgroup of index $2$.)
Then ${\rm Fix}_K(\phi(1)) = \langle y \rangle$, which is not normal in $G$.
A: *

*This happens quite often in the context of matrix groups. For instance, consider the group of matrices
$$M(n,(x_{ij}))=\begin{pmatrix}p^n & x_{12} & x_{13} & x_{14}\\ 0 & 1 & x_{23} & x_{24}\\ 0 & 0 & p^n & x_{34} \\ 0 & 0 & 0 & 1\end{pmatrix}$$
with $n\in\mathbf{Z}$, $x_{ij}\in\mathbf{Z}[1/p]$. This group is finitely generated, the normal subgroup $N$ of those $M(0,x,y,z)$ is not finitely generated (it has $\mathbf{Z}[1/p]$ as quotient). One can choose $\mathbf{Z}$ as the cyclic subgroup generated by $M(1,0,0,0)$. Then its fixed points on $N$ consists of those $M(0,(x_{ij}))$ for which $x_{ij}=0$ for $j-i\neq 2$ (i.e. only $x_{13}$ and $x_{24}$ can be nonzero). This is not a normal subgroup (it's not normal in the nontrivial nilpotent group $N$, since it has trivial intersection with the center).


*Another example: consider the group $\mathbf{Z}^2\ast \mathbf{Z}$ with presentation $\langle x,y,z\mid [y,z]\rangle$. Let $N$ be the kernel of the homomorphism onto $\mathbf{Z}$ mapping $y\mapsto 1$, $x,z\mapsto 0$. Then $N$ is free on $z$ and the $x_n=y^nxy^{-n}$ for $n\in\mathbf{Z}$. The group of fixed points of $y$ on $N$ is its free factor $\langle z\rangle$, which is not normal in $N$. And $N$ is not finitely generated.

(Even if you don't see why you have this description, $z$ is a fixed point of the $\langle y\rangle$-action but $xzx^{-1}$ is not, so the group of fixed points is not normal.)
