Large isometry groups of Kaehler manifolds

Question

Let $M$ be a closed simply-connected Kaehler manifold that is not isomorphic to a product of lower-dimensional Kaehler manifolds. Pick an orientation for $\mathbb{C}$; this endows $M$ with an orientation.

Assume that $M$ admits no self-diffeomorphisms (other than the identity) that preserve both metric and complex structure. This implies that the group of orientation-preserving isometries of $M$ is finite (this follows from the fact that every Killing vector field is holomorphic and the fact that isometry group is a compact Lie group in compact-open topology).

The question is: can this group have arbitrarily large order? What if we fix dimension of the manifolds?

Answer 1

A positive answer to this question should follow from Berger's holonomy classification and the following statement, which I believe is correct:

Statement. For any dimension $n$ there exists a positive constant $c(n)$ such that for any $m>c(n)$ the cyclic group $\mathbb Z_m$ can not act effectively by Kaehler isometries on any Hyper-Kaehler manifold of real dimension $4n$.

The argument works as follows. Consider the Holonomy group $H$ of $M$. Since $M$ is not a metric product, the holonomy should be irreducible. Since by our assumptions $M$ doesn't have Killing vector fields, it is not a symmetric space. So we can apply Berger's classification of holonomies: https://en.wikipedia.org/wiki/Holonomy#The_Berger_classification

Since our manifold is Kaehler, we deduce that the holonomy can be either $U(n)$, i.e. the manifold is just Kaehler, or it is $SU(n)$ and the manifold is a Calabi-Yau, or it is $Sp(n)$ and the manifold is Hyper-Kaehler. Now, it is easy to see that since $M$ has an isometry that doesn't preserve the complex structure, we can not be neither in the first nor in the second case. So $M$ is a Hyper-Keahler manifold with a Hyper-Keahler metric $g$.

Let $G$ be the finite group of isometries of $(M,g)$. Recall that $M$ has a three-dimensional space $\mathbb R^3$ of symplectic forms compatible with $g$ (corresponding to different complex structures). Clearly, $G$ is acting on $\mathbb R^3$ so we have a homomorphism $\varphi: G\to SO(3)$. By our assumptions the homomorphism has zero kernel (otherwise the elements in the kernel of $G$ would preserve the original complex structure). So $G$ is a finite subgroup of $SO(3)$. In case $|G|>60$ it is either diherdral or cyclic. Either way it contains a cyclic subgroup of index $2$. Let call this subgroup $G_0$. Then $G_0$ is fixing a vector in $\mathbb R^3$. So it also preserves the corresponding complex structure $J$ on $M$, i.e. it is acting by Kaehler authomorphisms on $(M,J,g)$. Hence if the Statement is correct we are done.

Remark. I have to admit that I don't know if the Statement is proven. It definitely holds for K3 surfaces, but since there is no classification of Hyper-Kaehler manifolds in higher dimensions, such a result might be harder to prove.