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 $\phi^\*[\omega\_N]=[\omega\_M]\in H^2(M, \mathbb{R})$?



Non-symplectomorphic isometric compact Kähler manifolds
The answer to your first question is 'no' and the answer to your second question is 'yes'.


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.


