Conjugacy of subgroups of free groups Let $F(X)$ be the free group on a finite set $X$. Suppose that $H \subseteq F(X)$ is a subgroup where $[F(X):H]$ is finite.
What can we say about the size of $[F(X):N(H)]$ where $N(H)$ is the normaliser of ? (Other than the trivial $[F(X):H] = [F(X):N(H)] [N(H):H]$)
That is, what can we say about the number of subgroups conjugate to $H$?
 A: A classical result (which can be seen topologically) asserts that if $|X|=r$ and $[F(X):H]=n$ then $H$ is a free group over $nr-(n-1)$ generators. It is quoted as a result of Schreier in this paper (p. 188) :
http://cms.math.ca/openaccess/cjm/v1/cjm1949v01.0187-0190.pdf
At the end (theorem 5.2) you have a theorem giving a formula for computing the number of subgroups of index $n$ in $F_r$.
It is far from what you are asking... As @Qiaochu Yuan said, it is hard to say better than the number of conjugates can be any divisor of $n$ so you cannot expect a better answer.
How does appear the group $H$? Do you have a set of free generators for it (in terms of the generators of $F_r$)? If you do, you might get a more accurate answer...
If you don't, as @Qiaochu Yuan said it is usually better to understand free groups and their finite index subgroups with topology. It boils down to give some wedge of circles covered with another wedge of circles. Some references for this :
https://www.math.cornell.edu/~hatcher/AT/ATch1.pdf
Combinatorial group theory, Lyndon and Schupp. 
