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4 HenrikRüping's answer gives conjugate solutions. I originally omitted to mention that I would quite like the solutions to be non-conjugate. I subsequently edited the question to include this.; added 29 characters in body

Let $F$ be the free group on (say) two generators, $a$ and $b$. Let $A$ and $B$ be (freely reduced) elements of $F$. Let $W(X, Y)$ denote a word on the words $X, Y$.

-Is it ever true that the equation $W(a, b) = W(A, B)$, has finitely many non-conjugate solutions ?(by conjugate solution I mean there exists a word $V$ such that $V^{-1}AV = A^{\prime}$ and $V^{-1}BV=B^{\prime}$)?

For example, take $W(a, b) = a^{-1}b^nab^m$. We therefore want to find $A$, $B$ such that $a^{-1}b^nab^m = A^{-1}B^nAB^m$, but we . We can take $A=b^ia$ and $B=b$ for all $i$ i$, and so this equation has infinitely many (non-conjugate) solutions. In fact,$a\mapsto b^ia$,$b \mapsto b$defines an automorphism of$F$(as free groups are Hopfian). Further, different$i$'s give different coset representatives of Out(F), and so a related question would be, -Does there exist a word$W \in F$such that there are only finitely many outer automorphisms$\phi$such that$W\phi = W$? I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example,$W(a, b) = a$. However, this doesn't quite work, as then$b$can be whatever we want (essentially, exclude this because its boring). Any help/ideas of papers to look at would be greatly appreciated. 3 deleted 14 characters in body; added 12 characters in body Let$F$be the free group on (say) two generators,$a$and$b$. Let$A$and$B$be (freely reduced) elements of$F$. Let$W(x, y)$W(X, Y)$ denote a word on the letters words $x, y$X, Y$. -Is it ever true that the equation$W(a, b) = W(A, B)$has finitely many solutions? For example, take$W(a, b) = a^{-1}b^nab^m$. We therefore want to find$A$,$B$such that$a^{-1}b^nab^m = A^{-1}B^nAB^m$, but we can take$A=b^ia$and$B=b$for all$i$and so this equation has infinitely many solutions. In fact,$a\mapsto b^ia$,$b \mapsto b$defines an automorphism of$F$(as free groups are Hopfian), and so a related question would be, -Does there exist a word$W \in F$such that there are only finitely many automorphisms$\phi$such that$W\phi = W$? I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example,$W(a, b) = a$. However, this doesn't quite work, as then$b$can be whatever we want (essentially, exclude this because its boring). Any help/ideas of papers to look at would be greatly appreciated. 2 added 2 characters in body Let$F$be the free group on (say) two generators,$a$and$b$. Let$A$and$B$be (freely reduced) elements of$F$. Let$W(x, y)$denote a word on the words letters$x, y$. -Is it ever true that the equation$W(a, b) = W(A, B)$has finitely many solutions? For example, take$W(a, b) = a^{-1}b^nab^m$. We therefore want to find$A$,$B$such that$a^{-1}b^nab^m = A^{-1}B^nAB^m$, but we can take$A=b^ia$and$B=b$for all$i$and so this equation has infinitely many solutions. In fact,$a\mapsto b^ia$,$b \mapsto b$defines an automorphism of$F$(as free groups are Hopfian), and so a related question would be, -Does there exist a word$W \in F$such that there are only finitely many automorphisms$\phi$such that$W\phi = W$? I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example,$W(a, b) = a$. However, this doesn't quite work, as then$b\$ can be whatever we want (essentially, exclude this because its boring).

Any help/ideas of papers to look at would be greatly appreciated.

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