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 $\omega_1$. But, in fact, because $\omega_1$ and $\omega_2$ are both $g$-parallel, their difference is $g$-parallel and hence $g$-harmonic. But, by the Hodge theorem, if their difference is exact, then it must be zero. Thus, they must be equal and hence we can simply use the identity map to align them.

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.