Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$? Is there a similarity between $\Bbb Z \times \Bbb Q$ and $\Bbb Z \times \Bbb N$? (EDITED: Both of these sets have anti lexicographic order, that is $(a,b)<(c,d)⟺b<d$ or $(b=d$ and $a<c)$).
Definition of similarity: Let $(A, <)$ and $(B,≺)$ be partially ordered sets. Then $f:A\rightarrow B$ is a similarity between $A$ and $B$ if $f$ is bijective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$ and $\forall m,n \in B, m≺n \Rightarrow f^{-1}(m)<f^{-1}(n)$. We write $A \simeq B$ if $A$ and $B$ are similar.
We usually check if the sets are (not) similar by checking for possible similarity invariants. (Let $A,B$ be posets. Property P is similarity invariant if $A \simeq B \Rightarrow (P(A) \iff P(B))$. That is, if $P(A)$ and $\lnot P(B)$, then $A \not \simeq B$.)
I have tried to find an invariant that one set has and the other doesn't, but haven't been able to find it. For example, they are both not separable (they both don't have a countable subset that's dense in them), they both aren't complete (there exists a nonempty subset that is bounded from above but doesn't have a supremum ( $\Bbb Z \times \{0\}$ is example for both), they both aren't locally finite, they are both unbounded on both sides, they both aren't dense in themselves.
On other hand, I can't find a similarity between them but I also don't know how to prove that there is no similarity between them.
(Also, if $(A, <)$ and $(B,≺)$ are both totally ordered (and in my case, they are), then $f:A\rightarrow B$ is similarity between $A$ and $B$ if $f$ is surjective and $\forall x,y \in A, x<y \Rightarrow f(x)≺f(y)$.)
 A: First some terminology.
What you call a 'similarity' is usually called an isomorphism. A slightly easier way of formulating that property is $f: A \to B$ is an isomorphism if $f$ is bijective and for all $x, y \in A$ we have $x < y \Leftrightarrow f(x) < f(y)$.
A useful invariant will be the following.

An order $(L, <)$ is called a dense linear order if for any $x < y$ in $L$ there is $z \in L$ such that $x < z < y$.

A useful fact about dense linear orders is that every interval in them is again a dense linear order. More precisely, if $L$ is a dense linear order then for any $a < b$ in $L$ we have that $\{x \in L : a < x < b \}$ is a dense linear order.

In $\mathbb{Z} \times \mathbb{Q}$ there is a copy of $\mathbb{Q}$, namely $Q = \{(0, x) : x \in \mathbb{Q}\}$. Clearly this is a dense linear order.
I claim that there is no subset of $A \subseteq \mathbb{Z} \times \mathbb{N}$ that is a dense linear order. Suppose for a contradiction that there is such an $A$. Then for some $n \in \mathbb{N}$ we must have that $A \cap (\mathbb{Z} \times \{n\})$ contains at least two elements $a$ and $b$, because otherwise $A$ would be isomorphic to $\mathbb{N}$. Without loss of generality we assume $a < b$. The interval $\{ x \in A : a < x < b \}$ can then not be a dense linear order because it is a subset of (something isomorphic to) $\mathbb{Z}$, so we find our contradiction.

I will also leave my old answer, which was assuming that we order the sets with the usual lexicographical ordering (instead of the anti lexicographical ordering).
The order $\mathbb{Z} \times \mathbb{Q}$ a dense linear order, but $\mathbb{Z} \times \mathbb{N}$ is not. For example, there is no element strictly between $(0, 5)$ and $(0, 6)$ in $\mathbb{Z} \times \mathbb{N}$.
A: Your sets are not just partially ordered, of course, they are totally ordered (and total ordering is an invariant of partially ordered sets).
In any totally ordered set $(S,<)$, for any two elements $r,s$, let's say that $s$ is a successor of $r$ if $r < s$ and there does not exist $t \in S$ such that $r < t < s$.
In $\mathbb Z \times \mathbb N$, every element $(m,n)$ has a successor, namely $(m+1,n)$. And the property "every element has a successor" is an invariant property. But in $\mathbb Z \times \mathbb Q$ it is not true that every element has a successor, in fact no element has a successor.
