When does the doubling map preserve measure ? Let $(G,+)$ be a locally compact (abelian) group endowed with its natural Haar measure. The doubling-map $T_2 : g \in G \mapsto g + g $ defines an endomorphism of $G$.

Is there some (natural) condition on $G$ which ensures that the map $T_2$ is measure-preserving?

When $G$ is finite, it is of course the same as asking $G$ to be of odd order (i.e. that there's no $2$-torsion). But $T_2$ may fail even when $G$ has no $2$-torsion, as shows $G = \mathbb{R}$. The group of $2$-adic integers $\mathbb{Z}_2$ (with its natural topology) shows that lack of 2-torsion and compactness are not sufficient either.
It is well-known that $T_2$ is measure-preserving when $G = \mathbb{T} = \mathbb{R}/ \mathbb{Z}$, or even when $G = \mathbb{T}^d$ (despite the $2$-torsion). I believe this is still true for any compact connected (abelian) group (I think connectedness would prevent pathologies such as $\mathbb{Z}_2$) but I have no convincing argument so far.
 A: For an endomorphism $\rho$ to preserve Haar measure on a compact group $G$, it is necessary and sufficient that $\rho$ be surjective.  One can easily check that the image $\rho^*(m)$ of Haar measure $m$ under $\rho$ is a translation-invariant probability measure as long as $\rho$ is surjective.  If $\rho$ is not surjective, the image $H=\rho(G)$ is a compact subgroup, either of positive finite index or infinite index.  In either case, $\rho^{-1}(H)$ will not have the same Haar measure as $H.$
When $G$ is compact and connected, the doubling map is surjective (one way to see this is to observe that the dual $\widehat{G}$ is torsion-free, so the doubling map induces an injective map on the dual).
More generally, if the dual of the compact group $G$ lacks $2$-torsion, then the doubling map will be surjective, and so will preserve Haar measure. 
Edit: The following argument and conclusion are in error, as Yves' comment shows.

The structure theorem for locally compact abelian groups reduces the general question to two cases: (1) $G$ is compact, and (2) $G$ is discrete. Fix a compact group $G$ where the doubling map preserves Haar measure, and let $\rho:G\to G$ be the doubling map.  By the structure theorem, $G$ has an open subgroup $W$ isomorphic to $K \times \mathbb R^n$ for some nonnegative integer $n,$ where $K$ is a compact group.  Since the doubling map does not preserve Haar measure on $\mathbb R$, we must have $n=0.$  If $\rho$ preserves Haar measure on $G$ then the induced map $\tilde{\rho}:G/W\to G/W$ preserves Haar measure on $G/W,$ so $\tilde{\rho}$ is injective.
Since $\tilde{\rho}$ is injective, we have $\rho^{-1}(W)=W.$  Thus $\rho(W)$ is a subgroup of $W$, and by the argument in the first paragraph, we necessarily have $\rho(W)=W.$
We have (edit: not) shown:
The doubling map on G preserves Haar measure if and only if $G$ has a compact open subgroup $W$ such that:
(i) The doubling map on $G/W$ is injective (i.e. $G/W$ lacks 2-torsion).
(ii)  $\widehat{W}$ lacks $2$-torsion.
