(I will use additive notation. In particular, $n\mid x$ for $n\in\mathbb N$ and $x\in G$ means there exists $y$ such that $ny=x$.)
It follows from Szmielew’s quantifier elimination that two elements $x,y$ of an abelian group $G$ satisfy the same first-order formulas if and only if $$m\mid nx\iff m\mid ny$$ for all integers $n,m\ge0$; this reduces to the following two special cases:
- $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
- $p^l\mid p^kx\iff p^l\mid p^ky$ for all primes $p$ and $l>k\ge0$ (where we may assume that $p^k$ divides the order of $x$ and/or $y$).
Now, if $G$ is finite, this is equivalent to the existence of an automorphism $\phi$ such that $\phi(x)=y$, because elementary equivalent finite structures are isomorphic. Moreover, in a finite abelian group, condition 2 implies condition 1.
(I’m sure there is a purely algebraic proof of this using the classification of finite abelian groups and the like, but well, I’m a logician, not an algebraist.)