Macpherson in *a survey of homogeneous structures*, states that there are many $\aleph\_0$-categorical structures which are not homogeneous. I'd like to know more of such examples.


**Edit:** The homogeneity mentioned here is the ultrahomogeneity that is defined as every isomorphism between two finite substructures of a structure $M$ can be extended to an automorphism of $M$. There is another homogeneity known as $ℵ\_0$-homogeneity that is defined as two $n$-tuples with the same type in $M^n$ must be on the same orbit of $\rm Aut$$(M)$.


$\aleph\_0$-categorical structures need not be ultrahomogeneous, but are always $\aleph\_0$-homogeneous. So $\aleph\_0$-homogeneous is a weaker notation than ultrahomogeneous in general. These two types of homogeneity become equivalent if and only if $\rm Th$$(M)$ has quantifier elimination.



Examples of $\aleph_0$-categorical nonhomogeneous structures
Here is my favorite example.



> 
> **Theorem.** Fix $n\geq 1$. Then there is a unique (up to isometry) countable metric space $(M\_n,d)$ satisfying the following properties:
> 
> 
> 1. $d(x,y)\in\{0,1,2,\ldots,n\}$ for all $x,y\in M\_n$
> 2. Any finite metric space with distances in $\{0,1,2,\ldots,n\}$ embeds as a subspace of $M\_n$.
> 3. Any partial isometry between two finite subspaces of $M\_n$ extends to a total isometry of $M\_n$.
> 
> 
> 


I don't really know who to attribute this to (perhaps [Casanovas & Wagner](https://www.ams.org/journals/proc/2004-132-05/S0002-9939-03-07193-4/S0002-9939-03-07193-4.pdf), or [Delhomme, Laflamme, Pouzet, Sauer](https://arxiv.org/pdf/math/0510254.pdf)). The point is that the class of finite metric spaces with distances in $\{0,1,\ldots,n\}$ is a Fraisse class in an appropriate relational language, and so $M\_n$ is the Fraisse limit. In particular, for all $k\leq n$, add a binary relation $d\_k(x,y)$ interpreted as "$d(x,y)\leq k$". In this language, $M\_n$ is homogeneous. But....


View $M\_n$ as a graph under just the relation $d\_1(x,y)$. Then $M\_n$ is still $\aleph\_0$-categorical because the metric is "definable" from the graph language using existential quantifiers. (Specifically, the defining properties of $M\_n$ force the metric $d$ to coincide with the "path metric" given by distance $1$. E.g, $d(x,y)\leq 2$ iff $\exists z(d\_1(x,z)\wedge d\_1(z,y))$.) However, in the graph language, $M\_n$ is homogeneous if and only if $n=1$ or $n=2$. Indeed, if $n\geq 3$ then we can find points $a,b,c,d\in M\_n$ such that $d(a,b)=2$ and $d(c,d)=3$. In the graph language, the finite substructures $\{a,b\}$ and $\{c,d\}$ are isomorphic. But an automorphism of $M\_n$ has to respect the metric, and so there is no automorphism sending $(a,b)$ to $(c,d)$.


The point, of course, is that by looking only at the distance $1$ relation, we lose quantifier elimination.


Note, on the other hand, $M\_1$ is a countably infinite complete graph and $M\_2$ is the countable Rado graph, both of which are homogeneous as graphs.


By the way, $M\_n$ is an example of what is called a *metrically homogeneous graph*. For more on these, see [Cherlin's work](https://sites.math.rutgers.edu/%7Echerlin/Paper/inprep.html) on the classification program for these graphs.


