Let $F$ be a (finitely generated) free group, $H \leq F$ of infinite index. Is it possible that $$ \bigcup\_{g \in F} gHg^{-1} = F?$$



Union of conjugates in free groups
The problem of characterizing groups that are union of conjugates of a proper subgroup was considered in some papers by Wiegold and others.
In particular, you can look at 


[*Transitive groups with fixed point free permutations*](http://link.springer.com/article/10.1007%2FBF01224701), Archiv der Mathematik **27** (1976), 473-475.


[*Groups covered by conjugated of proper subgroups*](http://www.sciencedirect.com/science/article/pii/S0021869305001420), Journal of Algebra **293** (2005), 261–268.


It turns out that the answer to your question is **yes**. 


In fact, already the free group on two generators $F\_2 = \langle a, b \rangle$ can be covered by the conjugates of one of its proper subgroups. This is shown in the first paper linked above, see Example 3.1 page 474.


