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Is there a $K3$ surface (algebraic, complex) that has infinitely many automorphisms of finite order?

Many K3 surfaces have infinite automorphism groups. In particular, all K3 surfaces of Picard rank 20 have infinite automorphism groups. However, the above question is unclear to me.

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    $\begingroup$ Does conjugating a finite order automorphism by every element of an infinite automorphism group not do the trick? $\endgroup$
    – Will Sawin
    Commented Jun 2, 2022 at 15:15
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    $\begingroup$ Cojugating by a commuting element does not give a new automorphim. It is unclear to me that conjugating shall give infinitely many different automorphisms. $\endgroup$
    – Basics
    Commented Jun 2, 2022 at 15:34
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    $\begingroup$ To give a concrete example: a hypersurface of degree $(2,2,2)$ in $\mathbb{P}^1\times \mathbb{P}^1\times \mathbb{P}^1$ contains a subgroup $G$ isomorphic to the free product of 3 groups of order 2 , hence containing infinitely many elements of order 2. $\endgroup$
    – abx
    Commented Jun 2, 2022 at 15:35
  • $\begingroup$ Great! Thanks for the quick answer! $\endgroup$
    – Basics
    Commented Jun 2, 2022 at 15:43

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