Existence of automorphism taking a word to its inverse

Question

Let $F_2$ be free group on free generators $\{x,y\}$. We know that the inverse of $[x,y] = xyx^{-1}y^{-1}$ is $[y,x]$ and if we take automorphism $\phi$ generated by $\phi(x)=y, \phi(y)=x$ then $\phi$ takes the word $[x,y]$ to its inverse $[y,x]$. If we take the word $[x^2,y]$ then does there exist any automorphism or endomorphism of $F_2$ which takes the word $[x^2,y]$ to its inverse?

Answer 1

You can solve this problem by just playing around with the letters, changing things a little bit at a time.

The goal is to produce an isomorphism having the effect $$x^2 \, y \, x^{-2} \, y^{-1} \mapsto y \, x^2 \, y^{-1} \, x^{-2} $$ The first thing to notice is that the input word has an $x^2 y$ subword whereas the desired output word has an $x^2 y^{-1}$ subword. So we'll start by fixing up that $x^2 y$ subword using the automorphism $\psi(x)=x$, $\psi(y)=y^{-1}$ (which is its own inverse): $$x^2 \, y \, x^{-2} \, y^{-1} \xrightarrow{\psi} x^2 \, y^{-1} \, x^{-2} \, y $$ The next thing to notice is that this new word differs from the desired output word by just a cyclic permutation. But every cyclic permutation is induced by an inner automorphism, in this case the inner automorphism $i_y(g)=y \, g \, y^{-1}$: $$x^2 \, y^{-1} \, x^{-2} \, y \xrightarrow{i_y} y \, x^2 \, y^{-1} \, x^{-2} $$ So $i_y \circ \psi$ is an automorphism with the desired affect.

Answer 2

Yes, $x\mapsto y x y^{-1}$, $y \mapsto x y^{-1}$.