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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 $k \ne i.$

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    You are ignoring some Nielsen moves, no? For example, sending x_i to its inverse, or swapping two elements.2010-11-29
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    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.2010-11-29
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    aren't you also missing x1 -> x1x2 -> x1x2^-1 -> x1 (last one is right multiplication by x2)?2010-11-30

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