Let $G$ be a discrete locally compact group and let $\alpha: G \to G$ be an automorphism. Show that the module $\rm{mod}_{G}(\alpha)$ is 1.
In the case of a locally compact field $k$ and $\alpha$ being multiplication by an element $a$ with $\rm{mod}_{k}(a) < 1$, it is proven by considering the sequence $a^{n}$ which has $\rm{mod}_{k}(a^{n}) \to 0$, hence $a^{n} \to 0$, so $\{0\}$ is not open. For this reason, I was attempting to consider the sequence $\rm{mod}_{G}(\alpha^{n})$ with $\rm{mod}_{G}(\alpha) < 1$, but I wasn't sure what to conclude.
The Haar measure on a discrete group is counting measure, so $\mu(E)=\#E$ for any Borel set $E$ (where $\#E$ means $\infty$ if $E$ is infinite). It's clear that an automorphism $\alpha$ of $G$ preserves cardinality (in the naive since that it takes infinite sets to infinite sets and it takes sets of finite cardinality to sets of the same finite cardinality), so for any Borel set $E$, $\mu(\alpha(E))=\mu(E)$, hence the modulos of $\alpha$ is $1$.