Let $R$ denote the Rado graph, and let $c$ be a fixed vertex.

**Question 1.** Is the structure obtained by extending $R$ by the constant $c$ interpretable in $R$ *without* parameters?

By *interpretable* I mean first-order interpretable; see below for an equivalent formalism.

A related group-theoretic question is the following. Let $\text{Aut}(R)$ denote the group of automorphisms of the Rado graph.

**Question 2.** Is there an action of $\text{Aut}(R)$ on $R$ by automorphisms fixing $c$ which has finitely many orbits?

Edit (11 Oct'16):

Here's a purely group-theoretic weakening of Question 2:

**Question 2'.**
Is the group $\text{Aut}(R)$ isomorphic to a subgroup of $\text{Aut}(R,c)$,
the group of automorphisms of $R$ fixing $c$?

Edit: Question 2' has been answered positively by 'Rado the explorator' below. Unfortunately, this still doesn't answer Questions 1 and 2, since the induced action has infinitely many orbits.

Here is some background to this question, and some minor observations.

For a relational structure $\mathbb A$ and its element $c$, let $(\mathbb A,c)$ denote the structure obtained from $\mathbb A$ by adding the constant $c$.
If $\mathbb A$ has the property that for every constant $c$, $(\mathbb A,c)$ interprets without constants in $\mathbb A$, then let us say that $\mathbb A$ has *elimination of parameters*.
Equivalently, this means that any relational structure $S$ which interprets in $R$ with parameters also interprets in $S$ without parameters.

In a comment to this question: ω-categorical, ω-stable structure with trivial geometry not definable in the pure set I remarked that $(\mathbb N,=)$ has elimination of parameters, and stated that this holds for any $\omega$-categorical structure, but Alex Kruckman pointed to an error in my reasoning, but we figured out a proof for $(\mathbb N,=)$. Here is an even simpler proof.

**Fact.** $(\mathbb N,=)$ has elimination of parameters.

In the proof, I allow myself to use a more convenient syntax for defining interpretations in a structure $\mathbb A$, which I call *definable sets*. The syntax of sets *definable over* $\mathbb A$ allows using set-builder expressions with variables ranging $\mathbb A$, and which can be constrained using first-order formulas in the language of $\mathbb A$; additionally, we can take finite unions or tuples of such expressions, and nest these operations. For instance, the set $\{(x,y):x,y\in\mathbb N, x\neq y\}\cup\{x:x\in\mathbb N, x\neq 5\}\cup \{\{x,y\}:x,y\in\mathbb N\}$ is definable over $(\mathbb N,=)$, using the parameter $5$.
A relational structure $\mathbb B=(B,R_1,\ldots,R_n)$ is *definable over* $\mathbb A$ if $B$ and each relation $R_i$ is a set which is definable over $\mathbb A$. Up to isomorphism, structures definable over $\mathbb A$ are the same as structures which interpret in $\mathbb A$, using first-order interpretations (this correspondence preserves the used parameters), but sometimes using definable sets makes constructions easier, as e.g. below. When the definition does not involve parameters, we say that the structure is 0-definable.

**Proof**. We show that $(\mathbb N,=,c)$ is isomorphic to a structure which is definable over $(\mathbb N,=)$:
Indeed, the structure $(\mathbb N\cup\{\emptyset\},R_=,c)$,
where $c$ is the constant interpreted as $\emptyset\in\mathbb N\cup\{\emptyset\}$ and $R_==\{(m,m):m\in\mathbb N\}\cup\{(\emptyset,\emptyset)\}$ is clearly 0-definable over $(\mathbb N,=)$ and isomorphic to $(\mathbb N,=)$. (It is also easy to construct a two-dimensional interpretation in $(\mathbb N,=)$ without parameters).
$\square$

**Fact.** $(\mathbb Q,\le)$ has elimination of parameters.

**Proof**. Here the construction is slightly more interesting. A constant $c\in\mathbb Q$ splits $\mathbb Q$ into two parts: the rationals smaller than $c$ and the rationals larger than $c$, and the interaction between the two parts is very simple. The structure $(\mathbb Q,\le, c)$ is isomorphic to the following structure. It's domain is $X=\{(q,L):q\in\mathbb Q\}\cup\{(q,R):q\in\mathbb Q\}\cup\{C\}$, where $L,R,C$ are symbols, e.g. implemented as $\emptyset,\{\emptyset\},\{\{\emptyset\}\}$ which are 0-definable, and so is $X$. We now define the relation $\le$ on $X$ by $R_\le=\Big\{((q,L),(q',L)):q,q'\in \mathbb Q, q<q'\Big\}\cup \Big\{((q,R),(q',R)):q,q'\in \mathbb Q, q<q'\Big\}$$ $$\cup
\Big\{((q,L),(q',R)):q,q'\in \mathbb Q\Big\}\cup \Big\{((q,L),C):q\in \mathbb Q\Big\}\cup \Big\{(C,(q,R)):q\in \mathbb Q\Big\}$,

which is also clearly 0-definable over $(\mathbb Q,\le)$. Finally, $(\mathbb Q,\le,c)$ is isomorphic to $(X,R_\le,C)$.$\square$

The question of this post is the following.

**Question 1.** Does the Rado graph $R$ have elimination of parameters?

Observe that a constant $c$ in $R$ splits the Rado graph into two parts: the neighbors $A$ of $c$ and the non-neighbors $B$. Each part is isomorphic to the Rado graph, and the interaction between the two parts is like a random bipartite graph. The structure $(R,A,B)$ is isomorphic to a random graph with a random partition. Question 1 is therefore equivalent to the question whether the random graph with a random partition interprets in $R$ without parameters.

Another question is the following.

**Question 3.**
Which $\omega$-categorical structures have elimination of parameters?

In group-theoretic terms, if a structure $\mathbb A$ 0-interprets in $R$, then the group of automorphisms of $R$ acts continuously on $\mathbb A$, and this action is oligomorphic, i.e., it induces finitely many orbits on $\mathbb A$, on $\mathbb A^2$, on $\mathbb A^3$, etc. In particular, if $(R,c)$ interprets in $R$, then there is an action of $\text{Aut}(R)$ on $(R,c)$ by automorphisms, which has finitely many orbits. Question 2 asks whether such an action exists? (Continuity of this action follows from the fact that $\text{Aut}(R)$ has the small index property. Also, if the action has finitely many orbits on $R$, then it is oligomorphic).