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Let $F_n$ be the free group on $n$ generators, $n>1$.

If $\phi$ is a quasi-isometry (or a bijective bilipschitz equivalence) on $F_n$, then what can we say about the explicit form of $\phi$?

In particular, can we find a finite index subgroup $H<F_n$, some $\psi\in Aut(H)$ and some element $g\in F_n$ such that $\phi(h)=g\psi(h), \forall h\in H$? (Here I used the convention that $d(g_1, g_2)$=the word length of $g_1^{-1}g_2$.)

Any comments or references are appreciated!

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    $\begingroup$ As a metric space $F_n$ has a continuum of isometries, as they correspond to automorphisms of its Cayley graph, which is an infinite $2n$-regular tree. Therefore I do not think that any such description of isometries is possible. $\endgroup$
    – Ashot Minasyan
    Commented Dec 1, 2016 at 14:40
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    $\begingroup$ First, let's restrict ourselves only to isometries fixing the identity (there are still $2^\omega$ of them). But this means that $g=1$ for any such isometry. Next, let's show that any pair $(H,\psi)$, using your notation, has at most one preimage in the group of isometries of $F_n$ (fixing $1$). Indeed, if $\phi:F_n \to F_n$ is a non-trivial isometry fixing $1$, then it must interchange at least two branches $T_1$ and $T_2$, of the Cayley graph, which originate at some fixed vertex $v$. $\endgroup$
    – Ashot Minasyan
    Commented Dec 1, 2016 at 16:36
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    $\begingroup$ Now, $T_1$ is a connected subset of the Cayley tree, which has vertices arbitrarily far from $v$. Therefore $T_1$ must contain a vertex corresponding to an element $h \in H$ (because $H$ has finite index, so every element of $F_n$ is within a bounded distance from an element of $H$). Hence $\phi(h) \neq h$, as $\phi(h)$ is a vertex of $T_2$, which is disjoint from $T_1$. Thus $\phi$ cannot induce the identity on $H$, and the above claim follows. $\endgroup$
    – Ashot Minasyan
    Commented Dec 1, 2016 at 16:44
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    $\begingroup$ The group $G=PSL_2(\mathbf{Q}_p)$ is uncountable and QI to $F_n$. Its left translations are self-QIs and are pairwise non-equivalent (=not at finite distance). Conjugating by a QI $G\to F_n$ you get the same uncountable number of pairwise non-equivalent self-QIs on $F_n$. $\endgroup$
    – YCor
    Commented Dec 1, 2016 at 17:07
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    $\begingroup$ The question as stated is certainly not correct: a better question would be to ask whether $\phi$ is at most finite distance from some $h\mapsto g\psi(h)$. (When one defines the quasi-isometry group of a metric space, this is the notion of equivalence one uses.) As @YCor's example shows, however, the answer is still 'no'. $\endgroup$
    – HJRW
    Commented Dec 2, 2016 at 9:30

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