To show that the models are not isomorphic assume an isomorphism $f:\frak{A}\to\frak{B}$, take $(a_n)_{n\in\omega}$ be an increasing sequence such $a_i,a_j$ are from a different copy of $\Bbb Z$ for each $i\ne j$.

Now look at $f(a_i),f(a_j)$. If they are from the same copy of $\Bbb Z$ then there exists some natural number $n$ that is their difference, and so there is a first order formula that represent this ("There exists exactly $n-1$ different elements that are between $f(a_i)$ and $f(a_j)$), because $f$ is an isomorphism it would imply that there is exactly $n-1$ elements between $a_i,a_j$ which is a contradiction.

Now for each $a_i$ let $b_i$ be the unique $k\in\omega^*$ such that $f(a_i)∈{\Bbb Z}×\{k\}$, this result with $(b_i)_{i\in\omega}$ being an increasing sequence in $ω^*$, contradiction.

To see that they are elementary equivalent we want to look at the EF game of length $n$, $G_n(\frak A,B)$.

When the Spoiler pick an element from $ω$ of one of the structures, the Duplicator will choose the same element from $\omega$ in the other one.

Next if the Spoiler choose an element not from $\frak A\setminus\omega$, then the Duplicator will choose any element from ${\Bbb Z}×\{-n^n\}$ in $\frak B$, and dually if the Spoiler choose from $\frak B\setminus\omega$.

Next if the Spoiler chooses an element that is a finite distance from a chosen element, the Duplicator will just duplicate this distance with the corresponded element in the other model.

If the Spoiler chooses a new $\Bbb Z$, the Duplicator will duplicate the $\Bbb Z$ but will cap the distance between the index of the chosen $\Bbb Z$ to the index of existing chosen indexes to have $k^k$ where $k$ is the number of steps left to the game. This will leave enough room to full out any kind of structure the Spoiler will try to create.

To see that they are not atomic, let for each natural $n$ $φ_n(x)$ be "$x$ is at least $n$" and let $τ(x)$ be the partial type consistent of all $\phi_n$. Note that for each $a\in\frak A$ we have $a\notin ω⇔τ(a)$, in particular $τ⊆\operatorname{tp}(a)$.

Now let $ψ(x)$ be any formula (that does not contain $z\ne z$ for some variable $z$), $ψ(x)$ can only describe what kind of finite linear orders exists/don't exists bellow $x$ and what kind of finite linear orders above $x$ and the distance of those orders from $x$ if it is finite and their distance from $0$ if it is finite

We don't care about what happens above $x$ because the end segment of $a$ and the end segment of $b$ are isomorphic for all $a,b$.

Now take some $a\notin ω$ and assume that $\operatorname{tp}(a)$ is isolated by $ψ(x)$. Note that any finite linear order exists bellow $a$, in particular $ψ(x)$ only describe what kind of orders do exists bellow $x$ and the distance of those orders if it is finite.

Take a natural number $m$ that is bigger than the sum of the length of the orders that $ψ$ describe plus the distance from $x$ of those orders plus the distance from $0$ of those orders, and any element $b∈ω\setminus m$ will satisfy $ψ$, in particular $ψ(x)\;\not\!\!\!\implies φ_b(x)$ and hence does not isolate $\operatorname{tp}(a)$

To show that the models are not isomorphic assume an isomorphism $f:\frak{A}\to\frak{B}$, take $(a_n)_{n\in\omega}$ be an increasing sequence such $a_i,a_j$ are from a different copy of $\Bbb Z$ for each $i\ne j$.

Now look at $f(a_i),f(a_j)$. If they are from the same copy of $\Bbb Z$ then there exists some natural number $n$ that is their difference, and so there is a first order formula that represent this ("There exists exactly $n-1$ different elements that are between $f(a_i)$ and $f(a_j)$), because $f$ is an isomorphism it would imply that there is exactly $n-1$ elements between $a_i,a_j$ which is a contradiction.

Now for each $a_i$ let $b_i$ be the unique $k\in\omega^*$ such that $f(a_i)∈{\Bbb Z}×\{k\}$, this result with $(b_i)_{i\in\omega}$ being an increasing sequence in $ω^*$, contradiction.

To see that they are elementary equivalent we want to look at the EF game of length $n$, $G_n(\frak A,B)$.

When the Spoiler pick an element from $ω$ of one of the structures, the Duplicator will choose the same element from $\omega$ in the other one.

Next if the Spoiler choose an element not from $\frak A\setminus\omega$, then the Duplicator will choose any element from ${\Bbb Z}×\{-n^n\}$ in $\frak B$, and dually if the Spoiler choose from $\frak B\setminus\omega$.

Next if the Spoiler chooses an element that is a finite distance from a chosen element, the Duplicator will just duplicate this distance with the corresponded element in the other model.

If the Spoiler chooses a new $\Bbb Z$, the Duplicator will duplicate the $\Bbb Z$ but will cap the distance between the index of the chosen $\Bbb Z$ to the index of existing chosen indexes to have $k^k$ where $k$ is the number of steps left to the game. This will leave enough room to full out any kind of structure the Spoiler will try to create.

To see that they are not atomic, let for each natural $n$ $φ_n(x)$ be "$x$ is at least $n$" and let $τ(x)$ be the partial type consistent of all $φ_n$. Note that for each $a\in\frak A$ we have $a\notin ω⇔τ(a)$, in particular $τ⊆\operatorname{tp}(a)$.

Now let $ψ(x)$ be any formula (that does not contain $z\ne z$ for some variable $z$), $ψ(x)$ can only describe what kind of finite linear orders exists/don't exists bellow $x$ and what kind of finite linear orders above $x$ and the distance of those orders from $x$ if it is finite and their distance from $0$ if it is finite

We don't care about what happens above $x$ because the end segment of $a$ and the end segment of $b$ are isomorphic for all $a,b$.

Now take some $a\notin ω$ and assume that $\operatorname{tp}(a)$ is isolated by $ψ(x)$. Note that any finite linear order exists bellow $a$, in particular $ψ(x)$ only describe what kind of orders do exists bellow $x$ and the distance of those orders if it is finite.

Take a natural number $m$ that is bigger than the sum of the length of the orders that $ψ$ describe plus the distance from $x$ of those orders plus the distance from $0$ of those orders, and any element $b∈ω\setminus m$ will satisfy $ψ$, in particular $ψ(x)\;\not\!\!\!\implies φ_b(x)$ and hence does not isolate $\operatorname{tp}(a)$

Assume an isomorphism $f:\frak{A}\to\frak{B}$, take $(a_n)_{n\in\omega}$ be an increasing sequence such $a_i,a_j$ are from a different copy of $\Bbb Z$ for each $i\ne j$. Show that $f(a_i),f(a_j)$ also must be from a different copy of $\Bbb Z$ and construct an infinite strictly increasing sequence in $\omega^*$ to reach contradiction

To show that the models are not isomorphic assume an isomorphism $f:\frak{A}\to\frak{B}$, take $(a_n)_{n\in\omega}$ be an increasing sequence such $a_i,a_j$ are from a different copy of $\Bbb Z$ for each $i\ne j$.

Now look at $f(a_i),f(a_j)$. If they are from the same copy of $\Bbb Z$ then there exists some natural number $n$ that is their difference, and so there is a first order formula that represent this ("There exists exactly $n-1$ different elements that are between $f(a_i)$ and $f(a_j)$), because $f$ is an isomorphism it would imply that there is exactly $n-1$ elements between $a_i,a_j$ which is a contradiction.

Now for each $a_i$ let $b_i$ be the unique $k\in\omega^*$ such that $f(a_i)∈{\Bbb Z}×\{k\}$, this result with $(b_i)_{i\in\omega}$ being an increasing sequence in $ω^*$, contradiction.

To see that they are elementary equivalent we want to look at the EF game of length $n$, $G_n(\frak A,B)$.

When the Spoiler pick an element from $ω$ of one of the structures, the Duplicator will choose the same element from $\omega$ in the other one.

Next if the Spoiler choose an element not from $\frak A\setminus\omega$, then the Duplicator will choose any element from ${\Bbb Z}×\{-n^n\}$ in $\frak B$, and dually if the Spoiler choose from $\frak B\setminus\omega$.

Next if the Spoiler chooses an element that is a finite distance from a chosen element, the Duplicator will just duplicate this distance with the corresponded element in the other model.

If the Spoiler chooses a new $\Bbb Z$, the Duplicator will duplicate the $\Bbb Z$ but will cap the distance between the index of the chosen $\Bbb Z$ to the index of existing chosen indexes to have $k^k$ where $k$ is the number of steps left to the game. This will leave enough room to full out any kind of structure the Spoiler will try to create.

To see that they are not atomic, let for each natural $n$ $φ_n(x)$ be "$x$ is at least $n$" and let $τ(x)$ be the partial type consistent of all $\phi_n$. Note that for each $a\in\frak A$ we have $a\notin ω⇔τ(a)$, in particular $τ⊆\operatorname{tp}(a)$.

Now let $ψ(x)$ be any formula (that does not contain $z\ne z$ for some variable $z$), $ψ(x)$ can only describe what kind of finite linear orders exists/don't exists bellow $x$ and what kind of finite linear orders above $x$ and the distance of those orders from $x$ if it is finite and their distance from $0$ if it is finite

We don't care about what happens above $x$ because the end segment of $a$ and the end segment of $b$ are isomorphic for all $a,b$.

Now take some $a\notin ω$ and assume that $\operatorname{tp}(a)$ is isolated by $ψ(x)$. Note that any finite linear order exists bellow $a$, in particular $ψ(x)$ only describe what kind of orders do exists bellow $x$ and the distance of those orders if it is finite.

Take a natural number $m$ that is bigger than the sum of the length of the orders that $ψ$ describe plus the distance from $x$ of those orders plus the distance from $0$ of those orders, and any element $b∈ω\setminus m$ will satisfy $ψ$, in particular $ψ(x)\;\not\!\!\!\implies φ_b(x)$ and hence does not isolate $\operatorname{tp}(a)$