Let $F$ a free group acting freely on a graph $X$. Does this imply that $X$ is a tree?
If $F=\mathbb Z$, the answer is no. For example $\mathbb Z$ acts on $\mathbb Z^2$ by translations.
But what about the other free groups?
As $F_2$ contains $F_k$ as a subgroup $\forall k$, it would suffice to show that the claim is also falso for $F=F_2$.
However I could not think of any graph which is not a tree on which $F_2$ acts freely. In particular, letting one generetor ($b$) act trivially does not give a free action, because then $1\neq aba^-1$ acts as the identity.