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 $f^{*}K_{\overline{M}_{0,n}} = K_{\overline{M}_{0,n}}$.

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

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?