# Non-symplectomorphic isometric compact Kähler manifolds

Let $$(M, \omega_M, J_M)$$ and $$(N, \omega_N, J_N)$$ be compact Kähler manifolds. Denote $$g_M=\omega_M(\cdot, J_M\cdot)$$ and $$g_N=\omega_N(\cdot, J_N\cdot)$$.

Assume there is a diffeomorphism $$\phi:M\to N$$ such that $$\phi^*(g_N)=g_M$$. Is there a diffeomorphism $$\psi:M\to N$$ such that $$\psi^*(\omega_N)=\omega_M$$? What if we additionally assume that $$\phi^*[\omega_N]=[\omega_M]\in H^2(M, \mathbb{R})$$?

A simple example, when $$n\ge 2$$, is to let $$M = \mathbb{R}^{2n}/\Lambda$$ where $$\Lambda\subset \mathbb{R}^{2n}$$ is a lattice (i.e., a discrete, co-compact subgroup of $$\mathbb{R}^{2n}$$, and let $$g$$ be the (flat) translation-invariant metric on $$M$$. Then there are many translation-invariant $$g$$-orthogonal complex structures on $$M$$, parametrized by $$\mathrm{O}(2n)/\mathrm{U}(n)$$, a manifold of dimension $$n^2{-}n$$, and, for the generic pair $$J_1$$ and $$J_2$$ of such complex structures, there will not be a diffeomorphism of $$M$$ with itself that aligns the corresponding Kähler forms $$\omega_{J_1}$$ and $$\omega_{J_2}$$, for cohomological reasons.
A less trivial example is to let $$M$$ be a K3 surface with its Ricci-flat Kähler metric $$g$$. Then there is a $$2$$-sphere of $$g$$-orthogonal, $$g$$-parallel complex structures on $$M$$, and, for the generic pair of such structures, there will not be a diffeomorphism of $$M$$ with itself that aligns the corresponding Kähler forms.
For the second question, you might as well replace $$(N,g_N, J_N)$$ with $$(M, \phi^*(g_N),\phi^*(J_N))$$ so as to have $$N=M$$ and $$\phi$$ equal to the identity. Now you are asking whether, if $$\omega_1$$ and $$\omega_2$$ are two $$g$$-parallel $$2$$-forms on $$M$$ that are cohomologous, does it necessarily follow that there is a diffeomorphism of $$M$$ with itself that pulls $$\omega_2$$ back to $$\omega_1$$.
Well, because $$\omega_1$$ and $$\omega_2$$ are both $$g$$-parallel, their difference is $$g$$-parallel and hence $$g$$-harmonic. Then, by the Hodge theorem, since their difference is exact and $$g$$-harmonic, it must be zero. Thus, they must be equal, i.e., we can simply use the identity map to align them.