indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Hrushovski construction?
generic predicate?
In the same way that in nonstandard analysis one passes to elementary extensions of the field $\mathbb{R}$ which realizes more types (in particular the types of infinite and infinitesimal numbers), one can extend the standard model $(\mathbb{N}, 0, 1, +, \times, S)$ of first-order Peano arithmetic to proper elementary extensions that have infinite natural numbers.
These “points at infinity” that live in the nonstandard part tend to embody the uniform behavior of the numbers from the standard part. (For example, by compactness, the twin prime conjecture is true if and only if in some nonstandard model of arithmetic there exists just a single pair of nonstandard twin primes, and similarly Dirichlet’s theorem on arithmetic progressions can be reformulated as saying that for every coprime positive pair of numbers $a$ and $b$ there exists just one nonstandard prime congruent to $a$ modulo $b$.)
A nonstandard model of arithmetic is a proper elementary extension of the standard model $(\mathbb{N}, 0, 1, +, \times, S)$ of first-order Peano arithmetic.
This is all discussed in e.g. (Gitman).
While the standard model of arithmetic has no automorphisms, there exist countable models of arithmetic with continuum-many automorphisms.
A theorem due to Harvey Friedman states that every nonstandard model of arithemetic is isomorphic to some initial segment (in terms of the ordering discussed below) of itself.
A theorem due to Stanley Tennenbaum states that there is no countable nonstandard model of arithmetic for which there is an algorithm to compute addition or multiplication on the nonstandard part.
The usual ordering on the natural numbers is definable in PA since $a \leq b \iff \exists c$ such that $a + c = b$.
As remarked above, the usual ordering on $\mathbb{N}$ is definable in $\mathsf{PA}$. Since the theory defines it, every nonstandard model of arithmetic is linearly ordered. For every nonstandard number $\omega$, one can generate using addition and subtraction a copy of $\mathbb{Z}$ around $\omega$, like so:
Since addition respects the ordering, $\omega + \omega$ is greater than anything in this copy of $\mathbb{Z}$, and since multiplication respects the ordering also, $\omega \times \omega$ is greater than $n \times \omega$ for any finite $n$.
Similarly, if $\omega_1 < \omega_2$, then the copy of $\mathbb{Z}$ around $\omega_1$ is below the copy of $\mathbb{Z}$ around $\omega_2$ in the ordering.
It turns out that if you quotient out these copies of $\mathbb{Z}$ in the nonstandard part of (any) nonstandard model of arithmetic, the resulting order on the nonstandard part is dense and without endpoints.
Let’s prove this. We’ll give two proofs; they’re really the same proof, but one uses the language of an ultrapower (the “hypernatural numbers”) while the other doesn’t.
The ordering of the copies of $\mathbb{Z}$ in the nonstandard part of a nonstandard model of arithmetic is dense and without endpoints.
(First proof.)
The theory of arithmetic proves that every other element is divisible by $2$, so pick even representatives from every copy of $\mathbb{Z}$ in the nonstandard part. Let $a$ represent a copy of $\mathbb{Z}$ below a copy of $\mathbb{Z}$ represented by $b$. The theory of arithmetic also proves that when $a$ and $b$ are even,
and since we have that $\models a + n < b$ for each finite $n$, we can also see that
for each finite $n$ (resp. < $b - n$ for all finite $n$). This order is dense and has no endpoints, since for any copy of $\mathbb{Z}$ in the nonstandard part we can take an even representative and divide by $2$ to obtain another nonstandard number which is below this copy of $\mathbb{Z}$ (to see it has no upper bound, we can just look at $\omega, \omega^2, \omega^3, \dots$.)
(Second proof.)
Let $^*\mathbb{N}$ be any ultrapower nonstandard model of arithmetic. This is an uncountable and (at least) $\aleph_{1}$-saturated model of $\mathsf{PA}$, and its elements are equivalence classes (germs) of sequences $\overline{x}$ of natural numbers. The nonstandard part still consists of copies of $\mathbb{Z}$. Let $\overline{a}$ and $\overline{b}$ be nonstandard numbers such that $\overline{a}$ belongs to a lower copy of $\mathbb{Z}$ than $\overline{b}$ does. By the Los ultraproduct theorem, $\overline{a}$ and $\overline{b}$ must therefore differ as sequences $\mathcal{U}$-often, with an unbounded sequence of differences. We can assume $\overline{a}$ and $\overline{b}$ are sequences of even numbers, since any perturbation of $\overline{a}$ and $\overline{b}$ by a sequence of $0$s and $1$s doesn’t change what copy of $\mathbb{Z}$ $\overline{a}$ and $\overline{b}$ are in. Then the sequence $\overline{c}$ of their pointwise average will be between them, and will have unbounded $\mathcal{U}$-often difference from $\overline{a}$ and $\overline{b}$.
Again by the Los ultraproduct theorem, the arithmetic operations in $^*\mathbb{N}$ are computed pointwise, so this argument will work in any elementary submodel of $^*\mathbb{N}$. In particular, any nonstandard model $M \models \mathsf{PA}$ embeds diagonally into its ultrapower $M^{\mathcal{U}}$.
Peano arithmetic is the quintessential unstable theory, and so there are plenty of countable nonstandard models of Peano arithmetic ($2^{\aleph_0}$-many.) One can see this directly by omitting the types of nonstandard numbers whose only finite prime divisors is an arbitrary set of finite primes.
By the discussion above, the nonstandard part is linearly ordered and can be partitioned into copies of $\mathbb{Z}$, and in fact these copies of $\mathbb{Z}$ are densely ordered without endpoints.
Since the theory of dense linear orders without endpoints has the unique countable model $(\mathbb{Q}, <)$, this means that for any countable nonstandard model of arithmetic, the order type of the nonstandard part is $\mathbb{Q} \mathbb{Z}$, i.e. $\mathbb{Q}$, but every time there’s a rational number, you have a copy of the integers instead.
So in a way, the wild divergence among the countable models of $\mathsf{PA}$ comes very concretely from the arithmetic part of $\mathsf{PA}$, and not at all from the ordering. However, this niceness (“categoricity”) of the ordering on nonstandard models of arithmetic is also being controlled by the arithmetic part of $\mathsf{PA}$ (since that was all we used in the proofs of denseness above): if we were looking at just elementary extensions of $(\mathbb{N}, <)$, you would need to stipulate $\omega$-saturation to force the order type on the nonstandard part to be $\mathbb{Q} \mathbb{Z}$, because otherwise there would be nothing wrong with just adjoining a single point at infinity and generating a single copy of $\mathbb{Z}$ (possible with just the order because you can ask for the next or previous element in the ordering) around that for a countable elementary extension of $(\mathbb{N}, <)$.
Victoria Gitman, An introduction to nonstandard models of arithmetic
R. Kossak and J. H. Schmerl, The structure of models of Peano arithmetic, The Clarendon Press, Oxford University Press, Oxford, 2006, vol. 50.