I cannot seem to find any results at all on characteristic subgroup growth, even of free groups (and even of $F\_2$). By contrast, the growth function of all subgroups of finite index is well-understood, as is the growth function of all normal subgroups of finite index (it's the same as enumerating finite groups). The growth function for normal subgroups clearly provides an upper bound, and one can get a truly ghastly lower bound by, eg, intersecting all subgroups of a given index (eg), but I find it hard to believe one can't do better.


For free groups characteristic subgroups are *verbal*, but it is not clear how this helps: given a collection of "words" computing the index of the corresponding verbal subgroup seems very hard (not recursively computable?).



Counting characteristic subgroups
**4 Jan 2011: Edited to fix discussion of verbal subgroups**


I think $F\_2$ is expected to behave differently from higher free groups. For a finite simple group $G$, I think it's expected (known?) that all epimorphisms $F\_n \rightarrow G$ are equivalent up to $Aut(F\_n)$ when
$n > 2$, but there are many orbits for $n = 2$: in particular, the isomorphism class (orbit under the automorphism group of $G$) of the image of the commutator of the generators is an invariant. 


Epimorphisms $F\_n \rightarrow G$ are equivlaent to $n$-tuples of generators of elements of $G$ that generate. An $n$-tuple that does not generate $G$ must be contained in a proper subgroup; for a finite simple group, there are not many maximal subgroups of small index, so the vast majority of $n$-tuples actually generate.
If there's only one orbit, the orbit size is rougly $|G|^n/|Aut(G)|$; for nonabelian finite simple groups, the outer automorphism group is small, so the orbit size is $|G|^{(n-1)}$ up to a small constant. 


The corresponding characteristic subgroup is the kernel of the
homomorphism $F\_n \rightarrow \prod G\_i$, where the product is over all epimorphisms to $G$ in the orbit, up to automorphism of $G$. A subgroup of the product of two groups is the kernel of the map of the product to a common quotient; using this, since the factors are simple, this homomorphism is an epimorphism. Therefore,
the characteristic subgroup has huge index,
$|G|^{|G|^{(n-1)/a}}$ for some not large $a$ (about 2 or 3).


Even for $n = 2$, the orbits are fairly large, so the corresponding characteristic subgroup have huge index.


*I.e.* characteristic subgroups with nonabelian simple composition factors are very sparse. Characteristic subgroups with polycyclic factor group are the vast majority, for several reasons: there are a lot of polycyclic groups, they tend to have large automorphism groups,
and different polycyclic quotients of a free group tend to correlate, so the homomorphism to a product of a number of polycyclic quotients is often far from surjective. Thus,
these characteristic subgroups do not have such a huge index. I suspect that in fact characteristic subgroups with abelian quotient should dominate. If so, the answer is boring, since
such subgroups correspond 1-1 with characteristic subgroups of $Z^n$. This may be why there's not much literature on it.



> 
> Verbal Subgroups and Fully Invariant Subgroups
> 
> 
> 


A word $w$ in letters $X\_1, \dots, X\_k$ defines a function $f\_w: G^k \rightarrow G$, when $k$-tuples of elements of $G$ are substituted for the $X\_i$. As explained in comments, a *verbal* subgroup $H\_V \subset G$ is one that is generated by the images of $f\_w$, for $w$ in some set $V$.
In other words, $G/H\_V$ is what you get by the identities for $w \in V$ that 
$\forall (X\_1, \dots X\_k)\; \; w = 1$ (as well as the relations of $G$ if $G$ is not free).


A *fully invariant* subgroup $H$ of a group $G$ is one such that every endomorphism
of $G$ takes $H$ to a subgroup of $H$. Fully invariant subgroups are characteristic, but not necessarily vice versa. 


In comments, Andy Putnam recalled that fully invariant subgroups of a free group are the ones that are supposed to be verbal, and he was dubious that characteristic subgroups of free groups are verbal. Here is an example to show he is correct:
that characteristic subgroups of $F\_2$ are not necessarily verbal:


We'll start from two epimorphisms $f\_1$ from $F\_2$ to $A\_5$ that are in different orbits of the automorphism group of $F\_2$: let $f1$ send the generators go to the 5-cycle
$a = (1,2,3,4,5)$ and the order 2 element $b = (1,2)(3,4)$. The commutator of $a$ and $b$ is a 5-cycle.
Let $f\_2$ send the generators to $a$ and to the 3-cycle $c = (3,4,5)$. The commutator of $a$ and $c$ is a 3-cycle. The conjugacy class of the commutator of generators is invariant under automorphisms of $F\_2$, so the maximal characteristic subgroups $H\_1$ and $H\_2$ containing the kernels of $f\_1$ and $f\_2$ are different.


Now consider any word $w(X\_1, X\_2, \dots, X\_N)$ whose image is contained in $H\_1$.
Equivalently, in $F\_2/H\_1$, this word evaluates identically to $1$. This happens if
and only if the projection to each $A\_5$ factor evaluates identically to 1.
Therefore, it also evaluates identically to $1$ in $F\_2/H\_2$. Any verbal subgroup
contained in $H\_1$ is in fact contained in the intersection $H\_1 \cap H\_2$.


To get the actual maximal verbal subgroup contained in the kernel of an epimorphism
such as $f\_1$,
first, choose a representative for each orbit of the action of $Aut(A\_5)$ on
 the product of copies of $A\_5$ for each of the $60^2$
homomorphisms $F\_2 \rightarrow A\_5$, form the product of these homomorphisms,
 and let $P$ be the image.
$P$ comes equipped with a pair of generators; let $R$ be a set of relators. 
Then $R$ is a verbal description for the kernel of this map.


