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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 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.

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 $a^{-1}b^nab^m = A^{-1}B^nAB^m$. We can take $A=b^ia$ and $B=b$ for all $i$, 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.

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 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.

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 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.

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 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.

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 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.