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.