$\newcommand{\o}{\mathcal{O}}\newcommand{\g}{\mathcal{G}}$Let $\Bbb N$ not include $0$. I cite this text, exercise $3.11$.
First some definitions:
$(K;\varphi)$ is a topological dynamic system if $K$ is a nonempty compact Hausdorff space and $\varphi:K\to K$ a continuous map.
A point $x_0\in K$ is recurrent if for every open $\o\ni x_0$ there is $n\in\Bbb N$ with $\varphi^n(x_0)\in\o$. This is equivalent to infinite recurrence: the set $\{n\in\Bbb N:\varphi^n(x_0)\in\o\}$ is unbounded.
Given a compact group $(\g,\cdot)$ and a continuous map $\Phi:K\to\g$, the group extension of $(K;\varphi)$ by $\g$ along $\Phi$ is the topological dynamic system $(K\times\g;\psi)$ where $\psi:K\times\g\to K\times\g,\,(x,g)\mapsto(\varphi(x),\Phi(x)\cdot g)$ is the extension of the dynamics.
Important: If $x_0$ is recurrent in $K$, $(x_0,g)$ is recurrent in any group extension $(K\times\g;\psi)$ for all $g\in\g$.
I am supposed to show that:
Given a system $(K;\varphi)$, any recurrent $x\in K$ is again recurrent in $(K;\varphi^m(x))$ for all $m\in\Bbb N$.
The hint:
Use a group extension by the cyclic group $\Bbb Z_m:=\Bbb Z/m\Bbb Z$.
I am totally at a loss with this problem. I cannot associate a general $K$ to integers, so I am finding it very difficult to associate $\varphi^m$ with the extended dynamics.
To plot out the problem, I supposed I had already found a suitable $\Phi:K\to\Bbb Z_m$, to get the picture:
$$\psi(x,n)=(\varphi(x),\Phi(x)+n)$$
Hmm... nothing at all leaps out at me from this. We know $(x,n)$ is recurrent, but to associate this recurrence with recurrence in $(K;\varphi^m)$ is difficult as $\Phi\circ\varphi^k$ is a meaningless integer as far as I'm concerned. We have:
$$\psi^m(x,n)=\left(\varphi^m(x),n+\sum_{k=0}^{m-1}\Phi(\varphi^k(x))\right)$$
In particular I can say if we let $\Phi$ be a constant map, due to cyclicity, that:
$$\forall n\in\Bbb Z:\psi^m(x,n)=(\varphi^m(x),n)$$
Which is maybe useful... I am unclear on how to proceed. I think this is the only way in which I can leverage cyclicity.
I toyed with the idea of performing multiple group extensions, but again - without a good sense of what $\Phi$ should be, this is not sparking any further ideas.
$\newcommand{\o}{\mathcal{O}}$I could not find the result anywhere online, so I will post my response here for any future benefit.
What I did in the question was essentially correct. The finishing touch is to let $\Phi(x)=1$ for all $x$, and to note that $\Bbb Z_m$ has the discrete topology. Letting $\lambda\in\Bbb Z_m$ be arbitrary, if $n$ is divisible by $m$ (so $n=mk$ for some natural $k$): $$\psi^n(x,\lambda)=((\varphi^m)^k(x),\lambda)$$ But whenever $n$ is indivisible by $m$: $$\psi^n(x,\lambda)=(\varphi^n(x),\lambda+n)$$And $\lambda+n\neq\lambda$ in $\Bbb Z_m$.
The recurrence of $(x,\lambda)$ under $\psi$ implies all product neighbourhoods of the point contain a member of its orbit: pick an arbitrary open $\o\ni x$ and note that $\o\times\{\lambda\}$ is an open neighbourhood in the extended system of $(x,\lambda)$, so $\psi^n(x,\lambda)\in\o\times\{\lambda\}$ for infinitely many $n$, but for any $n$ not divisible by $m$ the integer coordinate is $\lambda+n\notin\{\lambda\}$. Hence all the recurrences under $\psi$ (which we know exist by an earlier result of the textbook) must bring multiples of $m$ into the neighbourhood $\o\times\{\lambda\}$. Thus infinitely many $n=mk$ have $(\varphi^m)^k(x)\in\o$, for arbitrary $\o$, so $x$ is recurrent in the system $(K;\varphi^m)$.