Elementary substructure and automorphism There is a useful condition to show that $ \mathbf{A} $ is an elementary substructure of $ \mathbf{B} $: If for all finite subset $ P \subset A $ and element $ b \in B $, there is an automorphism $ f $ of $ \mathbf{B} $ that fixes $ P $ pointwise and sends $ b $ into $ A $, then $ \mathbf{A} \preceq \mathbf{B} $. However, the converse is not true, but I am unable to find a counterexample. Any hint?
 A: Note that your condition can't possibly be satisfied if $A\not=B$ and $B$ is rigid, that is has no nontrivial automorphisms. So a counterexample to the converse would be provided by any proper elementary substructure of a rigid structure. 
Now, some natural rigid structures have no proper elementary substructures - e.g. $(\mathbb{N};<)$ or $(\mathbb{Q}; +,\times, <)$. But there are rigid structures with proper elementary substructures: via the downward Lowenheim-Skolem theorem, any uncountable structure in a countable language (for example) has lots of proper elementary substructures. So now our task is just:

Find an example of an uncountable rigid structure in a countable language.

There are several of these:


*

*Show that if $(L; <)$ is a well-ordering, then it is rigid. Can you find an uncountable well-ordering? (This is a very hard problem if you haven't seen some set theory already, but if you have this is a great one.)

*Show that any Archimedean ordered field is rigid. (HINT: Think about how $\mathbb{Q}$ sits inside such a field.) Can you think of an uncountable Archimedean ordered field?

*Given a set $X$, let $L_X$ be the language consisting of a unary relation symbol for each $x\in X$ and $S_X$ be the structure with domain $\mathcal{P}(X)$ (that is, an element of $S_X$ is exactly a subset of $X$) and interpretation given by $U_x^{S_X}=\{A\in S_X: x\in A\}$. Is $S_X$ rigid? How big is $S_X$ when $X$ (and hence the language $L_X$) is countably infinite?
