Coding finite sets in abelian groups

Question

Let $(G,\cdot, e)$ be an abelian group and let $T=Th(G,\cdot, e)$. I wonder if $T$ codes finite sets, that is: Let $U$ be the monster model of T. Given $X= \{x_{1},\dots,x_{n}\}$ there is some $d \in G$ such that for any automorphism $\sigma$, $\sigma(X)=X$ if and only if $\sigma(d)=d$.

Answer 1

Not always. For example, let $p$ be prime and let $G$ be an abelian group of exponent $p$ (so we can view $G$ as a vector space over $\mathbb{F}_p$). Let $v,w\in G$ be linearly independent, and let $V = \text{Span}(v,w)$, a $2$-dimensional subspace of $G$ with basis $\{v,w\}$. If $d\in G^n$ codes the pair $\{v,w\}$, then each component $d_i\in \text{dcl}(v,w) = V$. But the map swapping $v$ and $w$ extends to an automorphism of $G$ which fixes the pair $\{v,w\}$ and hence fixes each component $d_i$. So $d_i$ is a vector $(a,a)$ in $\{v,w\}$-coordinates, i.e. $d_i = av + aw\in \text{Span}(v+w)$. But $v\notin \text{Span}(v+w)$, so there is an automorphism $\sigma$ of $G$ which fixes $v+w$ and such that $\sigma(v)\neq v$ and $\sigma(v)\neq w$. This automorphism does not fix the pair $\{v,w\}$, but it does fix $d$, contradiction.

The argument above was easy because of the extreme homogeneity provided by linear algebra. It could be that some less homogeneous abelian groups do manage to code finite sets - I'm not sure.