In addition to the Jordan-Holder theorem for groups, there are various Jordan-Holder Theorems for other categories: 

1. Finite dimensional representations have filtrations whose associated graded consists of irreducible representations. Any other such associated graded is the same up to permutation of its elements.

2. Artinian modules have filtrations whose associated graded consists of simple modules. Any other such associated graded is the same up to permutation of its elements.

3. Finitely generated $A$-modules have filtrations whose associated graded consists of modules of the form $A / p$ for a prime $p$. Any other such associated graded is the same up to permutation of its elements.

There is a commonality to the proofs of these as well. I am wondering if someone has come up with a categorical version of the Jordan-Holder theorem, which in a sense encompasses these ones.

For instance, one might try to look at the category of subquotients $\text{SubQuot}(X)$ of an object $X$ in a category $C$. Objects in this category are pairs $(Z, Y)$ with $Y \in \text{Quot}(X)$ and $Z \in \text{Sub}(Y)$. Morphisms $f : (Y, Z) \rightarrow (Y', Z')$ are pairs of maps $Y \twoheadrightarrow Y'$ in $\text{Quot}(X)$ and $Z' \rightarrow \text{im}(Z \rightarrow Y \rightarrow Y')$ in $\text{Sub}(Y')$.

Say a filtration is then a sequences of subquotients $(Y_i, Z_i)$ where $0 \rightarrow Z_i \rightarrow Y_i \rightarrow Y_{i+1} \rightarrow 0$ is exact. Say a simple object is one without nontrivial quotient objects.

We can represent a pair $(Z, Y)$ with $(Z_X, Y)$, where $Z_X$ is the pullback of $Z \rightarrow Y$ along $X \rightarrow Y$. To construct the coproduct $(Y, Z) \amalg (Y', Z')$ of $(Y, Z)$ and $(Y', Z')$, we simply take the pushout $Y''$ of $Y$ and $Y'$ in $C$ (coproduct in $\text{Quot}(X)$), and the pullback $Z''_X$ of $Z_X$ and $Z'_X$ in $C$ (product in $\text{Sub}(X)$). The coproduct is represented by the pair $(Y'', Z''_X)$.

Coproducts can be used to refine filtrations. The Jordan-Holder Theorem for modules then asserts that, for two filtrations $\{ (Z_i, Y_i) \}_{i = 1}^n$ and $\{ (Z_j', Y_j') \}_{j = 1}^m$ with simple (no nontrivial quotient objects) subquotients $Z_i$ and $Z_j'$, we can take a mutual refinement $\{ (Z_i, Y_i) \amalg (Z_j', Y_j') \}_{1 \leq i \leq n, 1 \leq j \leq m }$. The mutual refinement, after discarding its redundant elements, has the same subquotients as both $\{ (Z_i, Y_i) \}_{i = 1}^n$ and $\{ (Z_j', Y_j') \}_{j = 1}^m$, since a filtration by simple subquotients should have only the trivial refinements.

I think this works for groups and modules. However, I am particularly interested to see if anyone can make something like this work for the third example above, which seems harder, because I don't quite see how it fits in with the rest.