Existence of automorphism taking a word to its inverse

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?

• There is an algorithm called Whitehead's algorithm. en.wikipedia.org/wiki/Whitehead%27s_algorithm Commented May 3, 2022 at 11:36
• @SeanEberhard I have not tried Whitehead algorithm but it only gives us whether they are automorphic. What if there exists an endomorphism doing the job?
– Shri
Commented May 3, 2022 at 13:06

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