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.