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Let $\overline{M}_{0,n}$ be the usual Deligne-Mumford compactification of $M_{0,n}$ the moduli space of smooth $n$-pointed rational curves.

The canonical divisor $K_{\overline{M}_{0,n}}$ can be written as a combination of the irreducible components of the boundary of $\overline{M}_{0,n}$.

If $f:\overline{M}_{0,n}\rightarrow\overline{M}_{0,n}$ is an automorphisms then $f^{*}K_{\overline{M}_{0,n}} = K_{\overline{M}_{0,n}}$.

If $n\geq 5$, is this enough to conclude that $f$ has to map the boundary $\partial\overline{M}_{0,n}$ to itself?

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    $\begingroup$ The argument would not be correct as the canonical divisor is automatically preserved as a line bundle, but not as a divisor. Moreover, the statement is false in general. When $n=4$, $\bar{M}_{0,4}=\mathbb{P}^1$ has many automorphisms ! $\endgroup$ – Olivier Benoist Sep 24 '14 at 20:17
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If $n\geq 5$ then $Aut(\overline{M}_{0,n})\cong S_n$ (http://arxiv.org/abs/1006.0987). For instance $\overline{M}_{0,5}$ is a Del Pezzo surface of degree five. Its automorphism group is well known to be $S_5$ (http://arxiv.org/abs/math/0610595).

In particular, any automorphism of $\overline{M}_{0,5}$ preserves the boundary.

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