Why does every finite subgroup of $\mathrm{Aut}(F_n)$ acts on a graph of Euler characteristic $n-1$?

My question is the following:

In a paper I read that:

Any finite subgroup of $\mathrm{Aut}(F_n)$ can be realised as agroup of baspoint-preserving isometries of a graph of Euler characteristic $1-n$. Why is this fact true?

Thanks for help.

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 you might look at Karen Vogtmann's survey paper on $Out(F_n)$ (which also discusses $Aut(F_n)$). – Grumpy Parsnip Jan 16 '12 at 13:19