(It is possible that an answer to this question can be found in the literature, but I couldn't find anything after searching for about an hour.)


Let $G$ be a compact, totally disconnected, second countable group. (Equivalently, a profinite group: it is the inverse limit of finite groups.) It makes sense to talk about normal series: $\{e\} \lhd \cdots\cdots G\_2\lhd G\_1\lhd G$ where $G\_n$ is a closed subrgoup of $G$, $G\_n$ is normal in $G\_{n-1}$ and $G\_{n-1}/G\_n$ is finite and $\bigcap\_{n\in\mathbb{N}}G\_n=\{e\}$. This is the same as having an inverse limit expresion
$$G\to \cdots \to H\_2\to H\_1\to \{e\}$$
where each $H\_n$ is finite.


Call the normal series a composition series if each $G\_{n-1}/G\_n$ is simple. (In the inverse limit picture, this is the same as the kernel of each map $H\_n\to H\_{n-1}$ being simple.)


Are the simple factors appearing in a composition series unique up to permutation? The result is false without compactness, but I do not know if it is true with it.



Uniqueness of composition series for profinite groups
Yes, the composition factors are unique up to permutation, and this can be derived from the Jordan-Hölder theorem for finite groups.


If $G$ is second countable, it has a composition series: one can obtain this by starting with some countable basis of identity neighbourhoods (which may be taken to consist of open normal subgroups of $G$, since every open neighbourhood of the identity contains an open normal subgroup), taking intersections to form a descending chain of open normal subgroups with trivial intersection, and then refining the series. So let us assume that we have a composition series $(G\_n)$. The equivalence class of $(G\_n)$ is defined by the number of times, say $f(S)$, each isomorphism type $S$ of finite simple group appears as a factor $G\_{n-1}/G\_n$.


Let $S$ be a finite simple group. Then there are two possibilities: either $f(S)$ is a non-negative integer, or $f(S) = \aleph\_0$. If $f(S)$ is finite, then there is a finite discrete quotient $G/G\_n$ of $G$ such that some (hence any) composition series for $G/H$ has $f(S)$ simple factors isomorphic to $S$. Moreover, every finite discrete quotient of $G$ has at most $f(S)$ copies of $S$ in any composition series: This is clear for $G/G\_n$ for all $n$ assuming Jordan-Hölder for finite groups, and then we use compactness to show that any finite discrete quotient $G/H$ of $G$ is a quotient of $G/G\_n$. If $f(S)$ is infinite, then $G$ has finite quotients with composition series that have arbitrarily many factors isomorphic to $S$. Thus $f(S)$ is an invariant of $G$ as a topological group, independent of the choice of composition series $(G\_n)$, so all composition series are equivalent.


One can define transfinite descending composition series for arbitrary profinite groups. Here I am not sure about uniqueness: the difficulty is in counting the cardinality of times a composition factor appears (assuming it appears infinitely many times).


