Normal subgroups of automorphism group of relational structure Let $S$ be the set of all finite permutations of $\mathbb{N}$, i.e. they fix all but a finite set, and $A\subset S$ the set of all even permutations. 
Theorem. The normal subgroups of $S_\infty$ are exactly the following four: $\{\mathrm{id}\}\subset A\subset S\subset S_\infty$ (see on Wikipedia).
Now, consider a model $M$ with domain $\mathbb{N}$ and let $\mathrm{Aut}(M)$ be the group of  automorphisms of $M$. $\mathrm{Aut}(M)$ is a closed subgroup of $S_\infty$ and every closed subgroup of $S_\infty$ equals $\mathrm{Aut}(M)$, for some model $M$.

My question: Are there any (partial?) results generalizing the above  theorem to $\mathrm{Aut}(M)$? I.e. what are the normal subgroups of $\mathrm{Aut}(M)$, for various structures $M$? Equivalently, if $A$ is a closed subgroup of $S_\infty$, what are the normal subgroups of $A$?

 A: Not an answer, but the closed subgroups of $S_\infty$ are classified in this paper by Bergman and Shelah:
George M. Bergman and Saharon Shelah, MR 2280223 Closed subgroups of the infinite symmetric group, Algebra Universalis 55 (2006), no. 2-3, 137--173.
To answer Dima Pasechnik's question in the comments, the Baer-Schreier-Ulam theorem follows from the result proved by Manfred Droste and R Gobel, 1979.
CLOSER TO AN ANSWER
In their 2011 paper, MacPherson and Tent show (as conjectured in my comment) that these are often simple (the below is the beginning of the math review)
Dugald Macpherson and Katrin Tent, MR 2824528 Simplicity of some automorphism groups, J. Algebra 342 (2011), 40--52.:

Let $M$ be a countably infinite first-order relational structure and let $\mathrm{Aut}(M)$ be the automorphism group of $M$. The authors show that if $M$ is transitive, free, and homogeneous, and if $\mathrm{Aut}(M)$ is not the full symmetric group, then $\mathrm{Aut}(M)$ is simple. Transitive means that $\mathrm{Aut}(M)$ acts transitively on $M$; free means the class of finite substructures of $M$ has the free amalgamation property; and homogeneous means that isomorphisms between finite substructures extend to $\mathrm{Aut}(M)$.

A: Answer was edited
Here is a characterization I came up with:
Definition


*

*Let $G=Aut(M)$ (equiv. $G$ is a closed subgroup of $S_\infty$) and $N$ is a subgroup of $G$. Define $x\sim_N y$ iff there exists an automorphism $g\in N$ such that $g.x=y$.

*$G$ respects $\sim_N$ if for all $x,y$, $$x\sim_N y\text{ iff } g.x\sim_N g.y$$


Lemma Let $M$ be a countable model, $G=Aut(M)$ and $N$ be a closed subgroup of $G$. The following are equivalent:


*

*$N\vartriangleleft G$

*$G$ respects $\sim_N$


Proof. Unfold the definitions
 \begin{align*}
   N & \vartriangleleft G \text{ iff }\\
%  \forall g\in G\; \forall n\in N\; \exists n'\in N,\; & g^{-1}ng=n' \text{ iff }\\
 \forall x\in M \forall g\in G\; \forall n\in N\; \exists n'\in N,\; & 
g^{-1}ng.x=n'.x \text{ iff }\\
  \forall x,y \in M \forall g\in G,\; & (\exists n\in N,\; y=g^{-1}ng.x) 
\leftrightarrow  (\exists n'\in N,\; y=n'.x) \text{ iff }\\
 \forall x,y \in M \forall g\in G,\; & (\exists n\in N,\; g.y=ng.x) 
\leftrightarrow  (\exists n'\in N,\;y=n'.x) \text{ iff }\\
  \forall x,y \in M \forall g\in G,\; & 
(g.x\sim_N g.y) \leftrightarrow  (x\sim_N y)
\end{align*} 
End of Proof.
Stabilizers
As a special case we prove that if $X$ is a $G$-invariant set, then point-wise stabilizer of $X$ is a normal subgroup of $G$. 
Definition If $G$ is a group acting on $\mathbb{N}$ and $n\in\mathbb{N}$, then $G_n$ denotes the stabilizer of $n$ in $G$. If $X$ is a subset of $\mathbb{N}$, $G_{X}$ denotes the pointwise-stabilizer of $X$.
Lemma 1 For all $n,g$, $G_{g(n)}= g G_n g^{-1}$.
Proof. $f\in G_{g(n)}$ iff $f(g(n))=g(n)$ iff $g^{-1}(f(g(n)))=n$ iff $g^{-1}(f(g))\in G_n$ iff $f\in g G_n g^{-1}$. $\square$
Corollary 2 For $X\subset\mathbb{N}$, $G_{g(X)}=g G_X g^{-1}$.
Lemma 3 Let $G=Aut(M)$, for some $M$, and let $X$ be a $G$-invariant set. Then the pointwise-stabilizer of $X$ is a normal subgroup in $G$.
Proof.  Let $g\in G$. Since $X$ is G-invariant, $g(X)=X$. By Corollary 2, $G_X=G_{g(X)}=g G_X g^{-1}$, which proves the result. $\square$.
Some Comments


*

*By Lemma 3, the number of normal subgroups of $G$ correlates with the number of $G$-orbits. We can construct an example of $G$ with $\aleph_0$-many normal subgroups, even with $2^{\aleph_0}$-many normal subgroups. 

*In view of Lemma 3, $Aut(M)$ is a simple group, only if $Aut(M)$ acts trasitively on $M$, i.e. has only one orbit. So, it is necessary in the result mentioned by Igor Rivin that the $Aut(M)$ acts transitively on $M$.

*By a theorem of Dana Scott, every $Aut(M)$-orbit is defined by an $L_{\omega_1,\omega}$-sentence (in the same vocabulary as $M$). If $N$ is a normal subgroup of $Aut(M)$, what can we say about the action of $N$ on $M$?

