# Inverted pair of complex analytic families

I read the following "problem" in an old set of notes of Morrow and Kodaira which focused on deformations of complex manifolds:

Find a pair of complex analytic families $\lbrace M_t\rbrace$ and $\lbrace N_t\rbrace$ with $|t|<1$ such that $M_t=N_0$ for $t\ne 0$ and $N_t=M_0$ for $t\ne0$, with $M_0\ne N_0$. (Every manifold assumed compact.)

As of 1971 there were no known examples. Up to date, is this possibly true?

Naively it seems impossible. In the spirit of Lefschetz fibrations, we have $M_0$ as a critical level set amongst the $M_t$ (thinking of cylinders which get pinched via a vanishing cycle). It is then not generic, and any perturbation of this fiber (locally about the nodal point) would alter the topology. So to me the inverted family $\lbrace N_t\rbrace$ couldn't exist, because it would force a generic $M_{t_0}$ to take over the role as the critical level set amongst the $N_t$, but there is nothing "non-regular" about it.
If the manifolds are allowed to be noncompact, perhaps it's easy, again via a rough sketch: take $\lbrace M_t\rbrace$ to be the family of open cylinders which pinch the vanishing cycles at $t=0$, and $\lbrace N_t\rbrace$ to be the family of "pinched" cylinders where the pinching point pushes off to infinity as $|t|\to 0$.

Edit: The problem is more subtle than I thought, and my example with Lefschetz fibrations doesn't apply. From the definition of a "complex analytic family", all $M_t$ are diffeomorphic to each other (in particular, they're all nonsingular spaces), and so we can only distinguish them by their holomorphic structures. In other words, $M_0\ne N_0$ means diffeomorphic but not complex-analytically homeomorphic.

Suppose that $\mathcal M\to S$ is a miniversal deformation of $M_0$. Then the given family $\{M_t\}$ defines a morphism $\mu:\Delta\to S$, where $\Delta$ is a complex disc. Since $M_t=N_0\ne M_0$, the map $\mu$ is non-constant and $\mu(\Delta-\{0\})$ is contained in the stratum of $S$ that consists of points $s$ such that $\mathcal M_s$ is isomorphic to $N_0$. Now this stratum is smooth and its dimension, which we have just proved to be at least $1$, equals $\dim Aut_{M_0}-\dim Aut_{N_0}$; one reference for this, in the algebraic context (but in the current analytic context the result will be the same), is Theorem 2.7 of arXiv:1210.0342 by Hyland, Ekedahl and Shepherd-Barron. (In positive or mixed characteristic, where group schemes need not be reduced, this dimension becomes $\dim Lie(Aut_{M_0})-\dim Aut_{N_0}$.)
So $\dim Aut_{N_0}<\dim Aut_{M_0}$. From the assumptions, the opposite inequality is also valid, and we have a contradiction.