The connection between quantifier elimination, $\omega$-categorical and ultrahomogenous A relational structure $A$ with an $\omega-$categorical theory $Th(A)$ is ultrahomogenous iff $Th(A)$ admits quantifier elimination. (*)
I was wondering wether the structure $A$ has to be countable...
Think of the following example:
We have the set $A= (0,1)_\mathbb Q \cup (1,2)_\mathbb{R}$  and one binary relation $<$ (the linear order). 
I think $A$ admits quantifier elimination, is $\omega-$categorical (because the theory is that of a dense linear order) but NOT ultrahomogenous as $\mathbb{Q}$ and $\mathbb{R}$ are not isomorph.
What do you think?
Thank you in advance for any help!

Supplement
I haven’t had much time to do modeltheory the last weeks, but today I tried to proof the other direction of *. And I would appreciate some help again!
Claim: If a countable structure $A$ has an $\omega$-categorical theory and admits QE, then it is ultrahomogeneous.
Proof: Assume $Th(A)$ admits quantifier elimination, $\bar{a}, \bar{b} \in A^n$ and $ f: \bar{a} \mapsto \bar{b}$ is an isomorphism. We have to show: there exists an automorphism $f’$ so that $f'(\bar{a}) = \bar{b}$.
As $Th(A)$ is $\omega-$categorical and admits QE, there exists a quantifier-free formular $\varphi(\bar{x})$, that isolates tp($\bar{a}$). And because of the isomorphism $f$ we get: $A \models  \varphi( \bar{b})$, hence tp($\bar{a}$)=tp($\bar{b}$). At this point I don't know how to argue, that whenever two n-tuples have the same complete type, there exists an automorphism between them. Do you have an idea? Thank you!
 A: You're right, the theorem is only true with the additional hypothesis that $A$ is countable.
In your example, the structure $A$ is not ultrahomogeneous, because the $2$-element substructure $\{\frac{1}{3},\frac{2}{3}\}$ is isomorphic to the $2$-element substructure $\{\frac{4}{3},\frac{5}{3}\}$, but this isomorphism doesn't extend to an automorphism of $A$, since the interval $(\frac{1}{3},\frac{2}{3})$ is countable, while the interval $(\frac{4}{3},\frac{5}{3})$ is uncountable.

In response to your comment: The converse holds even for uncountable structures. However, the notion of ultrahomogeneity is really most relevant for countably infinite structures.
Claim: If a structure of any cardinality is ultrahomogeneous and has an $\aleph_0$-categorical theory, then it admits quantifier elimination.
Proof: Let $M$ be our ultrahomogeneous structure. It suffices to show that whenever two tuples $\overline{a}$ and $\overline{b}$ in any model of $T$ realize the same quantifier-free type, they realize the same complete type. Suppose not, so we have $\overline{a}$ and $\overline{b}$ in some model $N$ with $p(\overline{x}) = \text{tp}(\overline{a}) \neq q(\overline{x}) = \text{tp}(\overline{b})$, but $\text{tp}^\text{qf}(\overline{a}) = \text{tp}^\text{qf}(\overline{b})$. 
But $\aleph_0$-categoricity of $T$, there are only finitely many formulas in $n$ free variables up to equivalence, so every type is isolated (equivalent to a formula). Hence $p$ and $q$ are realized in $M$, say by $\overline{a}'$ and $\overline{b}'$. But now $\overline{a}'$ and $\overline{b}'$ realize the same quantifier-free type, so they generate isomorphic substructures of $M$, and by ultrahomogeneity, there is an automorphism of $M$ moving $\overline{a}'$ to $\overline{b}'$. Hence $\text{tp}(\overline{a}) = \text{tp}(\overline{b})$, contradiction.
