Theorem 1.1. A relation $R \subseteq M^n$ is definable if and only if every automorphism of every elementary extension of $M$ preserves $R$.
For a proof, the reader can see [4]. Suppose we want to use this theorem in the case of groups; we must assume that $G$ is a group, $S \subseteq G$ is an arbitrary subset, and $\mathcal{L}(S)$ is the extended language of groups with parameters from $S$. Clearly a singleton set $\{g\}$ is a unary relation in $G$, so we can restate the above theorem, for $S$-definability of elements in $G$.
Question: At theorem above: a singleton set $\{g\}$ can be regarded as a unary relation in $G$. Why?
Corollary 1.2. An element $g$ is $S$-definable in $G$ if and only if, for any elementary extension $G'$ of $G$ and every automorphism $\alpha:G' \rightarrow G'$, if $\alpha$ fixes elements of $S$, then it fixes also $g$.
If $G$ is a finite group, then the only elementary extension of $G$ is $G$ itself, because there exists a first order sentence which says that $G$ has $m$ elements ($m$ is the order of $G$), so any elementary extension of $G$ must have order $m$. Hence, for the case of groups, we have...
Question: If $G$ is a finite group then the only elementary extension of $G$ is $G$ itself. Why?
For your first question, a unary relation on $G$ is just a subset of $G$. So given $g\in G$, the singleton subset $\{g\}\subseteq G$ is a relation. Formally, we can add a new relation symbol $R$ to the language and interpret it as $R^G = \{g\}$, so that $G\models R(a)$ if and only if $a = g$.
Then to say that the relation is preserved by all automorphisms of elementary extensions of $G$ is to say that for all $G'\succeq G$, and all $\sigma\in \text{Aut}(G')$, $\sigma(g) = g$.
For the second question, see André Nicolas' comment.