# $\operatorname{Stab}(x_1)$ in $\text{Aut}( F(x_1,\ldots,x_n ) )$

Let $F_n$ be an $n$-generator free group with a free basis $x_1,\ldots,x_n.$ Is it true that the stabilizer of $x_1$ in $\mathrm{Aut}(F_n)$ is generated by all left and right Nielsen moves $\lambda_{ij}$ and $\rho_{ij}$ such that $i \ne 1$ and by the element of order two $\epsilon_n$ such that $\epsilon_n(x_n)=x_n^{-1}$ while other elements of the basis remain fixed.

Let $i \ne j$ and $1 \le i,j \le n.$ The left Nielsen move $\lambda_{ij}$ takes $x_i$ to $x_j x_i$ and the right Nielsen move $\rho_{ij}$ takes $x_i$ to $x_i x_j;$ both $\lambda_{ij}$ and $\rho_{ij}$ fix all $x_k$ with $k \ne i.$

-
You are ignoring some Nielsen moves, no? For example, sending x_i to its inverse, or swapping two elements. –  user641 Nov 29 '10 at 22:58
You are right. Subgroup generated by all Nielsen moves is of index two in Aut(F_n). Thus I need also an element of order two, say $\epsilon_n$ which inverts $x_n$ and fixes other $x_k.$ I am editing the question accordingly, thanks. –  krof Nov 29 '10 at 23:09
aren't you also missing x1 -> x1x2 -> x1x2^-1 -> x1 (last one is right multiplication by x2)? –  Alon Amit Nov 30 '10 at 5:10

Curiously, this question is a generalisation of this question which I asked a wee while ago (but long after you asked this one). I was asking about the 2-generator case, you are wanting to know about the $n$-generator case.
So, for the $2$-generator case, F(a, b), the stabiliser of $a$ consists of the automorphisms $\phi: a\mapsto a, b\mapsto a^iba^j$, and a proof (or two, or maybe three...) can be found in the question I linked too. This translates as,
$\phi\in\operatorname{Stab}(a)$ if and only if $\phi\in \langle\beta, \gamma, \beta^{\alpha}, \gamma^{\alpha}\rangle$, where $\gamma$ is the Dehn twist, $$\gamma: a\mapsto a, b\mapsto ba$$ $$\alpha: a\mapsto a^{-1}, b\mapsto b$$ $$\beta: a\mapsto a, b\mapsto b$$ and $g^h=g^{-1}hg$ (but here $\alpha^{-1}=\alpha$).