Suppose $e : \mathbb R \to F$ is an elementary embedding in the language of ordered fields. Can there exist an elementary embedding $e' : \mathbb R \to F$ such that $e \not= e'$? Note that it would have to be the case that for some undefinable $r \in \mathbb R$, $e(r) \not= e'(r)$, and $e(r) - e'(r)$ is infinitesimal.



Can nonstandard fields contain $\mathbb R$ in different ways?
Yes, this is possible. Let $F$ be a nonarchimedean strongly $\omega$-homogeneous real-closed field such that $\mathbb R\subseteq F$ (which exists by model-theoretic general reasons). Fix an infinitesimal element $\epsilon>0$. By quantifier elimination for real-closed fields, $\pi$ and $\pi+\epsilon$ have the same type (the only property of $\pi$ I need here is that it is transcendental), hence by homogeneity, there exists an automorphism $f$ of $F$ such that $f(\pi)=\pi+\epsilon$. Then $f\restriction\mathbb R$ is an (elementary) embedding of $\mathbb R$ in $F$ different from inclusion.


In fact, there is nothing special about $\mathbb R$:



> 
> **Proposition.** For any first-order structure $A$, the following are equivalent:
> 
> 
> 1. For every structure $B$, there is at most one elementary embedding of $A$ in $B$.
> 2. All elements of $A$ are definable.
> 
> 
> 


2${}\to{}$1 is obvious. For 1${}\to{}$2, if $a\in A$ is not definable, the type $\{x\ne a\}\cup\{\phi(x):A\models\phi(a)\}$ is consistent, hence there exists an elementary extension $B$ of $A$ and $b\in B$ such that $b\ne a$ and $(B,a)\equiv(B,b)$. By elementarily extending $B$ further if necessary, we may assume $B$ to be strongly $\omega$-homogeneous. Thus, there exists an automorphism $f$ of $B$ such that $f(a)=b$, and then $f\restriction A$ gives a second elementary embedding of $A$ in $B$.


