Let $a$ and $b$ be elements of group $G$ and assume that for any natural number $n$ and for any first-order formula $\varphi (x_1,x_2,...,x_n)$ with $n$ free variables in the language of groups, $G \models (\varphi (a,a, ... , a) \leftrightarrow \varphi (b,b,...,b))$
Is there always a group automorphism of $G$ that maps $a$ to $b$?
If $b$ is the image of $a$ under an automorphism of $G$, then $a$ and $b$ satisfy the same first order formulas. The question is whether the converse of that statement holds or not?
Here is a counterexample with a non-abelian group. I think the question in the abelian case is quite interesting.
Given an abelian group $(U,\cdot)$, define the group $H_U$ to be the semidirect product of $U$ with a cyclic group of order $2$ that acts on $U$ by sending an element to its inverse. So, explicitly, $H_U=\{ua_U^{\varepsilon}:u\in U,\varepsilon\in\{0,1\}\}$, where $a_U$ has order $2$ and $ua_U=a_Uu^{-1}$ for all $u\in U$. Note that, if $H$ has no elements of order $2$, then the centralizer of $a_U$ in $H_U$ is just $\{1,a_U\}$.
Let $(U,\cdot),(V,\cdot)$ be torsion-free divisible abelian groups of $\mathbb{Q}$-dimensions $1$ and $2$, respectively. Define $G=H_U\times H_V$. Then $(a_U,1)$ and $(1,a_V)$ are not conjugate by an automorphism of $G$, since $(a_U,1)$ has centralizer $\{1,a_U\}\times H_V$ and $(1,a_V)$ has centralizer $H_U\times\{1,a_V\}$, and these groups are not isomorphic for dimension reasons. However, I claim that $(a_U,1)$ and $(1,a_V)$ satisfy all the same formulas in the language of groups.
There are several ways to see this, but here is one argument. Let $V_1<V$ be a $1$-dimensional subspace. Consider $H_{V_1}$ as a subgroup of $H_V$ by identifying $a_{V_1}$ with $a_V$. Then $H_U\times H_{V_1}$ is an elementary substructure of $G$. (I will write an explanation of this tomorrow as I am going to sleep now.) Moreover, there is an isomorphism $f:H_U\to H_{V_1}$ mapping $a_U$ to $a_{V}$, and now $(g,h)\mapsto (f^{-1}(h),f(g))$ gives an automorphism of $H_U\times H_{V_1}$ taking $(a_U,1)$ to $(1,a_V)$, giving the desired result.
Okay, so why is $H_U\times H_{V_1}$ an elementary substructure of $H_U\times H_V$? First note the following. If $(U,\cdot)$ is an abelian group with at least $2$ elements, then $(H_U,\cdot)$ is uniformly interpretable in $(U,\cdot)$: take the quotient of $U\times U$ under the equivalence relation identifying any two elements of form $(u,v)$ and $(u,w)$, where $v,w\neq 1$, let $a_U$ be the equivalence class of $(1,v)$ for some (any) $v\neq 1$, and then equip it with the usual multiplication rules for semidirect product.
If follows that, if $V_1\preccurlyeq V_2$ are abelian groups, and $V_1$ is an elementary substructure of $V_2$, then $H_{V_1}$, which is naturally a subgroup of $H_{V_2}$ by identifying $a_{V_1}$ with $a_{V_2}$, is an elementary substructure of $H_{V_2}$.
Now, in the setting of the counterexample, $V_1$ is an elementary substructure of $V$ by quantifier elimination, and so $H_{V_1}$ is an elementary substructure of $H_V$. It follows that the two-sorted structure $((H_U,\cdot),(H_{V_1},\cdot))$, which no relations between the two sorts, is an elementary substructure of $((H_U,\cdot),(H_V,\cdot))$. Since $(H_U\times H_{V_1},\cdot)$ and $(H_U\times H_V,\cdot)$ are definable by the same formula in these two structures, respectively, the claim follows.