The following question is driving me bananas.  

I am given a split extension  

$0 \to \mathbb{Z}/n\mathbb{Z} \to (\mathbb{Z}/m\mathbb{Z})\ltimes (\mathbb{Z}/n\mathbb{Z})\to \mathbb{Z}/m\mathbb{Z}\to 0,$  

with $m,n>2$ natural numbers, and an element $x\in \mathbb{Z}/n\mathbb{Z}$ satisfying $x^m=1$ but $x^j\neq 1$ for $j < m$. Multiplication by $x$ corresponds to the conjugation action of $\mathbb{Z}/m\mathbb{Z}$. I know also that $ 1 - x $ is a unit. I inject $\mathbb{Z}/n\mathbb{Z}$ into $\mathbb{Z}/n^2\mathbb{Z}$ by mapping the generator $1\in \mathbb{Z}/n\mathbb{Z}$ to $1\in \mathbb{Z}/n^2\mathbb{Z}$, then $2\in \mathbb{Z}/n\mathbb{Z}$ to $2\in \mathbb{Z}/n^2\mathbb{Z}$, *etc.* Call this mapping ι.   




> 
> What is the order of $(\iota x)^m-1$ in $\mathbb{Z}/n^2\mathbb{Z}$? 
> 


Based on naive calculations and on an optimistic temprament, I'm tempted to guess that it is exactly $n$. A certain topological quantity which I'm working with is naturally an element of the ideal generated by $\frac{(\iota x)^m-1}{n}\bmod n$ in $\mathbb{Z}/n\mathbb{Z}$ (I think), and it would be so much more elegant if I knew this ideal to be all of $\mathbb{Z}/n\mathbb{Z}$.



Order in $\mathbb{Z}/n^2\mathbb{Z}$ of an $m$th root of unity in $\mathbb{Z}/n\mathbb{Z}$
Dror Speiser provides a counterexample in a comment:



> 
> $x=2$, $n=3511$, $m= \mathrm{ord}\_n(x)=1755= \mathrm{ord}\_{n^2}(x)$. See wikipedia for more on Wiefrich primes and generalizations.
> 


