Existence of an automorphism of $F_2$ having some certain property Let $F_2 = \langle a, b \rangle$, the free group of rank $2$. 
Is there an automorphism $\phi$ of $F_2$ such that $\phi(b) = b$, $\phi^2 = 1$ and 
$\phi(a) = wa^{-1}bw^{-1}$ for some $w \in F_2$?
First I checked the induced map on the abelianizations but it did not help. Then I expressed $w$ as a product of powers of $a$ and $b$. There seems to be too many cases to consider.
 A: This cannot happen. Consider the abelinisation. One route is to realise that taking $w=1$ is an automorphism, and thus you would need $w=b^i$ by Nielsen's old result, which says that if two automorphisms of $F_2$ act identically on the abelinisation then they are equal in $\operatorname{Out}(F_2)$ (note: this doesn't work in $F_n$ for $n>2$ because of Torelli). You can find Nielsen's Theorem in the book Combinatorial Group Theory by Magnus, Karrass and Solitar. The result can then be worked out as follows:
Take $w=b^i$. Then because $a\phi=b^ia^{-1}b^{1-i}$, $b\phi=b$ we have the following working. $$\begin{align*}
a\phi^{2}
&=(b\phi)^i(a\phi)^{-1}(b\phi)^{1-i}\\
&=b^i(b^ia^{-1}b^{1-i})^{-1}b^{1-i}\\
&=b^ib^{i-1}ab^{-i}b^{1-i}\\
&=b^{2i-1}ab^{1-2i}\\
\end{align*}$$
Therefore we require $i=1/2$, which cannot happen.
It is interesting to note that $\phi$ has order two on $\operatorname{Out}(F_2)$ but not in $\operatorname{Aut}(F_2)$. Indeed, every element of $\operatorname{Out}(F_2)\cong GL_2(\mathbb{Z})$ is contained in a subgroup isomorphic to $D_6$ or $D_4$, because $GL_2(\mathbb{Z})\cong D_6\ast_{D_2}D_4$ (the reference is B. Zimmermann, Finite groups of outer automorphisms of free groups, Glasgow Mathematical Journal, although I think there is a paper of Culler which does the same thing). Therefore, find these (unique-up-to-conjugacy) subgroups and you find all possible finite order elements of $\operatorname{Aut}(F_2)$. You can then test these to see which actually do have finite order in $\operatorname{Aut}(F_2)$. I did this once, but I have no idea where I wrote it down... It is a nice exercise though, and I would recommend you try it.
