Let $w$ be a word in a free group $F_2$ of two generators $x_1, x_2$ such that there does not exist any endomorphism of free group which takes $w$ to $w^{-1}$. Let $w_1, w_2$ be two words in the same free group such that they generate rank $2$ free subgroup. Let $\phi$ be an endomorphism of $F_2$ generated by sending $x_i$ to $w_i$. I want to prove that for the word $u = \phi(w)$, there does not exist any endomorphism of $F_2$ inverting $u$. 

I have tried but did not get anywhere. This is what I have tried:  Suppose there exists an endomorphism $\psi$ of $F_2$such that $\psi(u) = u^{-1}$ but $\psi \circ \phi (w) = \phi(w^{-1}).$ $\phi(F_2)$ has rank 2 and $\psi$ inverts $\phi(w).$ By Sela and Kharlampovich-Myasnikov, non abelian free group have the same elementary theory. And it may happen that $\psi$ may not be invariant for $\phi(F_2).$ Certainly rank of $\psi(F_2) \cap \phi(F_2)$ is either 1 or 2. In both cases, I am stuck.

Different methods are also appreciated. 
Thanks in advance!