Saturated Models of same Cardinality are Isomorphic

Theorem
Let $T$ be an $\mathcal{L}$-theory.

Let $\kappa$ be an infinite cardinal.

If $\mathcal{M}$ and $\mathcal{N}$ are saturated models of $T$ and the cardinality of the universes of $\mathcal{M}$ and $\mathcal{N}$ are both $\kappa$, then $\mathcal{M}$ and $\mathcal{N}$ are isomorphic.

Proof
The idea of the proof is that since the models are saturated, we can define an isomorphism $f:\mathcal{N}\to\mathcal{M}$ by picking the image $f(x)$ of an element $x\in \mathcal{N}$ as something which realizes the type that $f(x)$ would need to satisfy in $\mathcal{M}$. This is done using a (transfinite) recursive construction which extends an elementary map at each step. The proof is very similar to that of Saturated Implies Universal; in both proofs we attempt to construct a formula-preserving map. In this case, however, the map must be bijective between the universes of the models.

First, since $\mathcal{M}$ and $\mathcal{N}$ have cardinality $\kappa$, we can denote their elements by $m_\alpha$ and $n_\alpha$ respectively for ordinals $\alpha < \kappa$.

For each $\alpha < \kappa$, let $A_\alpha$ denote the domain of the $f_\alpha$ we define below. Note that we do not know in advance what the domains will be.

Define $f_0 = \emptyset$.
 * Base case $\alpha = 0$:

Note that $f_0$ is trivially an elementary  embedding from $A_0 = \emptyset$ into $\mathcal{M}$.

Let $\displaystyle f_\alpha = \bigcup_{\beta < \alpha} f_\beta$.
 * Limit ordinals $\alpha$, assuming $f_\beta:A_\beta \to \mathcal{M}$ is defined and elementary, and that $|A_\beta|<\kappa$ for all $\beta < \alpha$:

If $\phi$ is an $\mathcal{L}$-sentence with parameters from $A_\alpha$,  then since it involves only finitely many such parameters, they must all  be contained in some $A_\beta$ for $\beta < \alpha$. But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.

Note that $|A_\alpha| < \kappa$ by Cardinality of Infinite Union of Infinite Sets.

We need to extend $f_\beta$ to some $f_\alpha$ so that the domain includes $n_\beta$, the image includes $m_\beta$, and so that $\mathcal{L}$-formulas with parameters from $A_\alpha$ are preserved by $f_\alpha$.
 * Successor ordinals $\alpha = \beta + 1$, assuming $f_\beta:A_\beta \to \mathcal{M}$ is defined and elementary, and that $|A_\beta| < \kappa$:

First we add $n_\beta$ to the domain:

Consider the subset $p = \{\phi(v, f_\beta(\bar{a})):\bar{a}\text{ is a tuple from }A_\beta \text{ and }\mathcal{N}\models\phi(n_\beta,\bar{a})\}$ of the set of $\mathcal{L}$-formulas with one free variable and parameters  from the image $f_\beta (A_\beta)$ of $A_\beta$ under $f_\beta$.

The set $p$ is a $1$-type over the image $f_\beta (A_\beta)$ in $\mathcal{M}$.

Since $|A_\beta| < \kappa$ by the inductive hypothesis and since by assumption $\mathcal{M}$ is  $\kappa$-saturated, this means that $p$ is realized in $\mathcal{M}$ by  some element $b$.

Thus $f'_\alpha = f_\beta \cup \{(n_\beta,b)\}$ is an elementary embedding $A_\beta \cup \{n_\beta\} \to \mathcal{M}$.

Now we add $m_\beta$ to the image:

This is done similarly to the above.

Consider the subset $q = \{\phi(v, \bar{a}):\bar{a}\text{ is a tuple from }A_\beta \cup \{n_\beta\} \text{ and }\mathcal{M}\models \phi(m_\beta, f'_\alpha(\bar{a}))\}$ of the set of $\mathcal{L}$-formulas with one free variable and parameters from $A_\beta \cup \{n_\beta\}$.

The set $q$ is a $1$-type over $A_\beta \cup \{n_\beta\}$ in $\mathcal{N}$.

Since $|A_\beta| < \kappa$ by the inductive hypothesis and hence $|A_\beta \cup \{n_\beta\}|<\kappa$ as well, and since by assumption $\mathcal{N}$ is $\kappa$-saturated, this means that $q$ is realized in $\mathcal{N}$ by some element $c$.

Thus $f_\alpha = f'_\alpha \cup \{(c, m_\beta)\} = f_\beta \cup \{(n_\beta,b), (c, m_\beta)\}$ is an elementary embedding $A_\beta \cup \{n_\beta\} \to \mathcal{M}$ which includes $m_\beta$ in its range.

Finally define $\displaystyle f = \bigcup_{\alpha < \kappa} f_\alpha$.

The map $f$ is an elementary embedding $\mathcal{N}\to \mathcal{M}$ since $\displaystyle \bigcup_{\alpha < \kappa} A_\alpha = \mathcal{N}$, any finite set of parameters from $\mathcal{N}$ must belong to some single $A_\alpha$, and $f\restriction A_\alpha$ is elementary.

$f$ is onto $\mathcal{M}$ since we have constructed it so that $m_\alpha$ is in the image of $f_{\alpha+1}$.

So, $f$ is a bijection.

Since elementary embeddings are by definition $\mathcal{L}$-embeddings, this means that $f$ is an isomorphism by definition.