Let $w$ be a word in free group $F$ on finitely many generators. We will look at $w$ as word map on groups. It is clear that there exists an endomorphism $\phi$ of $F$ such that $\phi(w) = w^{-1}$ if and only if for every group $G$, the image of word map ($\mathrm{Im}(w)$) is closed with respect to inverse. My question is if $w$ has a property which is for every finite group $G$, if $g \in \mathrm{Im}(w)$ then $g^{-1} \in \mathrm{Im}(w)$, then does there exist an endomorphism $\theta$ of free group $F$ such that $\theta (w) = w^{-1}$.
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3$\begingroup$ Note that an intermediate property is the existence of an endomorphism of the profinite completion of $F$ mapping $w$ to its inverse. $\endgroup$– YCorCommented Sep 19, 2022 at 20:57
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$\begingroup$ @YCor thanks. The free group $F$ sits inside its profinite completion $\hat{F}$ then we need to see that an endomorphism taking the word to its inverse can be restricted to the free group (or may be obvious). It is also not obvious that there will exist such an endomorphism. $\endgroup$– ShriCommented Oct 5, 2022 at 13:38
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