When is the group-theoretic notion of algebraic closure idempotent? In model theory, the notion of algebraic closure can be defined in terms of formulas having only finitely many realizers.  Alternatively, it can be defined in terms of automorphisms:  $a \in M$ is in the ALGEBRAIC CLOSURE or ACL (apud Hodges) of $A \subseteq M$ iff $\mathrm{Aut}(M / A) \cdot a$ is finite.
If $M$ is sufficiently homogeneous (or atomic), then the two notions of algebraic closure coincide, and arguably this is often the case when one talks about the latter notion.  But I'm curious what happens if $M$ is not so homogeneous. Interestingly, the claim that ACL(-) is idempotent is suspiciously missing from Section 4.1 of Longer Hodges.
I believe that ACL(ACL($X$)) is just ACL($X$) for finite $X$ by induction, but I don't know if this is true in a general setting.  (In order to find a counterexample, $M$ must be neither too homogeneous or too rigid, and this eludes me at the moment. Also, Neumann's lemma may be useful here?) Is ACL idempotent in a general setting?  If not, what's a general assumption under which it is idempotent?
 A: Here is an example to show that you will not get idempotence in general.

We will define a $2$-sorted structure $M=(X,P)$ in the language consisting of a single binary relation $E$ on the $X$-sort and a single binary relation $\in$ from the $X$-sort to the $P$-sort. $X$ consists of countably many $E$-equivalence classes, one of every finite size: denote the class of size $n$ by $Y_n$. $P$ is the power set of $X$, and $\in$ is interpreted in the natural way. Now the automorphisms of $(X,P)$ are in natural bijection with the automorphisms of $X$, and the automorphisms of $X$ are precisely those bijections $X\to X$ which fix every $Y_n$ setwise. In particular, every element of $X$ has finite orbit under $\operatorname{Aut}(M/\varnothing)$, so that $X\subseteq\operatorname{ACL}(\varnothing)$, whence $\operatorname{ACL}(\operatorname{ACL}(\varnothing))$ is the whole structure. However, if $a\in P$ is a subset of $X$ containing a single element from every $Y_n$, then $a$ has infinite orbit under $\operatorname{Aut}(M/\varnothing)$ and hence does not lie in $\operatorname{ACL}(\varnothing)$. Hence this notion of algebraic closure is not idempotent.

Note that you can easily modify this example to get a $1$-sorted counterexample $N$; just let $N=X\sqcup P$ and interpret $P$ as a new $E$-equivalence class. Then the counterexample still goes through.
