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(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 $n,m\ge0$; in other words, $x$ and $y$ have the same order in $G/mG$. This reduces to the following two special cases:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ 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 (or: bounded exponent) abelian group, condition 2 implies condition 1.

In fact, this shows that your condition $$\def\p#1{\langle#1\rangle}G/\p x\simeq G/\p y$$ is already necessary and sufficient: if it holds, then for any $m\ge0$, $$G/(\p x+mG)\simeq(G/\p x)/m(G/\p x)\simeq(G/\p y)/m(G/\p y)\simeq G/(\p y+mG),$$ thus $|\p x+mG|=|\p y+mG|$, i.e., $x$ and $y$ have the same order in $G/mG$.

(I’m sure there is a direct algebraic proof of this using the classification of finite abelian groups and the like, but well, I’m a logician, not an algebraist.)

(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 $n,m\ge0$; in other words, $x$ and $y$ have the same order in $G/mG$. This reduces to the following two special cases:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ 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 (or: bounded exponent) abelian group, condition 2 implies condition 1. Thus, as a summary: for finite $G$, $x$ is automorphic to $y$ iff

$$\text{$x$ and $y$ have the same order in $G/p^lG$ for each prime power $p^l\mid\lvert G\rvert$.}$$

In fact, this criterion shows that your condition $$\def\p#1{\langle#1\rangle}G/\p x\simeq G/\p y$$ is already necessary and sufficient: if it holds, then for any $m\ge0$, $$G/(\p x+mG)\simeq(G/\p x)/m(G/\p x)\simeq(G/\p y)/m(G/\p y)\simeq G/(\p y+mG),$$ thus $|\p x+mG|=|\p y+mG|$, i.e., $x$ and $y$ have the same order in $G/mG$.

(I’m sure there is a direct algebraic proof of this using the classification of finite abelian groups and the like, but well, I’m a logician, not an algebraist.)

(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:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ 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.)

(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 $n,m\ge0$; in other words, $x$ and $y$ have the same order in $G/mG$. This reduces to the following two special cases:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ 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 (or: bounded exponent) abelian group, condition 2 implies condition 1.

In fact, this shows that your condition $$\def\p#1{\langle#1\rangle}G/\p x\simeq G/\p y$$ is already necessary and sufficient: if it holds, then for any $m\ge0$, $$G/(\p x+mG)\simeq(G/\p x)/m(G/\p x)\simeq(G/\p y)/m(G/\p y)\simeq G/(\p y+mG),$$ thus $|\p x+mG|=|\p y+mG|$, i.e., $x$ and $y$ have the same order in $G/mG$.

(I’m sure there is a direct algebraic proof of this using the classification of finite abelian groups and the like, but well, I’m a logician, not an algebraist.)

(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:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ for all primes $p$ and $l>k\ge0$ such that $p^k$ divides the order of $x$ (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 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.)

(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:

1. $x$ and $y$ have the same order (i.e., $nx=0\iff ny=0$ for all $n$).
2. $p^l\mid p^kx\iff p^l\mid p^ky$ 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.)