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 $\nu:M\to N$ such that $\nu^\*(\omega\_N)=\omega\_M$, there is a diffeomorphism $\phi:M\to N$ such that $\phi^\*(J\_N)=J\_M$ and there is an orientation-preserving diffeomorphism $\chi:M\to N$ such that $\chi^\*g\_N=g\_M$.


Is there a diffeomorphism $\psi:M\to N$ such that $\psi^\*(\omega\_N)=\omega\_M$ and $\psi^\*(J\_N)=J\_M$ (and hence also $\psi^\*(g\_N)=g\_M$)?



Non-isomorphic compact Kähler manifolds that are biholomorphic, symplectomorphic and isometric
The answer is 'no, not necessarily'.


Consider the following example: Let $M=N=\mathbb{CP}^2$, let $(\omega\_0,J\_0)$ be the standard Fubini-Study Kähler structure on $M$. Now let $f$ be an arbitrary, but '$C^2$-small' smooth function on $M$, so that $\omega\_0 + t\,\mathrm{i}\,\partial\bar\partial f$ is nondegenerate (and hence symplectic) for all $0\le t\le 1$.


Let $\omega\_M = \omega\_0 + \mathrm{i}\,\partial\bar\partial f$ and let $J\_M= J\_0$. Let $\omega\_N=-\omega\_M$ and let $J\_N=-J\_0$. Note that $g\_M=g\_N$, so $(M,g\_M)$ and $(N,g\_N)$ are isometric via the identity map.


When $f$ is chosen sufficiently generically, the isometry group of $g\_M$ will consist of *only* the identity, so suppose this.


Note that $(M,J\_M)$ and $(N, J\_N)$ are biholomorphic, since $\mathbb{CP}^2$ is biholomorphic to its conjugate complex manifold.


By Weinstein's theorem, since $\omega\_M$ and $\omega\_0$ are $C^0$-close and cohomologous, there is a symplectomorphism between $(M,\omega\_M)$ and $(M,\omega\_0)$. Similarly, there is a symplectomorphism between $(N,\omega\_N)$ and $(N,-\omega\_0)$. Since, as has already been noted, $(M,\omega\_0)$ and $(N,-\omega\_0)$ are symmplectomorphic, it follows that $(M,\omega\_M)$ and $(N,\omega\_N)$ are symplectomorphic.


However, when $f$ is chosen sufficiently generically, the only map $\psi:M\to N$ that aligns the metrics $g\_M$ and $g\_N$ is the identity, which is neither a biholomorphism nor a symplectomorphism.


**Remark:** Of course, I woke up this morning and was struck by the fact that there is an even simpler example: Let $g$ be any metric on $S^2=\mathbb{CP}^1$ whose isometry group is trivial, let $J$ be one of the two $g$-orthogonal complex structures on $S^2$, and let $\omega\_J$ be the associated area $2$-form on $S^2$. Let $(M,\omega\_M,J\_M)=(S^2,\omega\_J,J)$ and let $(N,\omega\_N,J\_N)=(S^2,-\omega\_J,-J)$.


Then $(M,\omega\_M,J\_M)$ and $(N,\omega\_N,J\_N)$ are biholomorphic, symplectomorphic, and isometric, but not isomorphic as Kähler manifolds.


Moreover, it is not difficult to choose an unramified rational curve in $\mathbb{CP}^2$ (it may have crossings, but that doesn't matter, for example, the generic irreducible cubic with one node but no cusp would certainly do) so that the metric on the normalized curve (which is an $S^2$ topologically) induced by the Fubini-Study metric on $\mathbb{CP}^2$ has no nontrivial isometries. Thus, one can even construct a pair of examples with the additional property requested by the OP that the metric be induced from the Fubini-Study metric by an immersion into $\mathbb{CP}^2$. (Using $\mathbb{CP}^3$ as a target, one could even arrange the map to be an embedding.)


