Really rigid varieties Are there (complex, projective) varieties $X_0$ with the following property?

Every (flat, say) family $\mathfrak X\to T$ over a reasonable and reasonably big class of bases with a fiber over a closed point equal to $X_0$ is a trivial family, in the sense that $\mathfrak X$ is $T\times X_0$.

If one asks this with $T$ restricted to local Artinian rings, say, one gets the class of (infinitesimally) rigid varieties, but I wonder if there are "globally rigid" examples.
Later: As Francesco and unknowngoogle observe, it is quite unreasonable to ask for such strong rigidity... :)
 A: If $X_0$ is a smooth projective variety over $\mathbb{C}$ such that $H^1(X_0, T_{X_0})=0$, then any deformation over a disk $D_{\epsilon}:=\{z \in \mathbb{C} | |z| < \epsilon \}$ is trivial for $\epsilon$ small enough. 
Notice that, for a general scheme,  $H^1(X_0, T_{X_0})=0$ only ensures that there are no infinitesimal deformations (i.e. deformations over Artin rings); however, since $X_0$ is projective one shows that nearby fibers in any flat family of projective schemes are isomorphic to $X_0$ [see Hartshorne, Deformation theory, p. 39].
For instance, the complex projective space $\mathbb{P}^n$ is globally rigid.
Another class of example is given by Fake Projective Planes: since they are $2$-dimensional ball quotients, they are globally rigid by Mostow Rigidity Theorem. In fact, their moduli space consists of $100$ distinct isolated points.
Let me also mention Beauville surfaces: they are surfaces $S$ of general type with $p_g=q=0$, $K_S^2=8$ (i.e. they are fake quadrics) obtained as $$S=(C_1 \times C_2)/G,$$
where $G$ is a finite group acting faitfully on the smooth curves $C_i$ and freely on their product, in such a way that $C_i/G \cong \mathbb{P}^1$ and  the two maps $ C_i \longrightarrow C_i/G$ are both branched at three points. 
Since three points on the projective line have no moduli, by general results of Catanese it follows that such surfaces are globally rigid, and in fact their moduli space is given again by a finite collection of reduced points.   
Finally, notice that the $T=D_{\epsilon}$ is probably one of the most raisonable choices for checking global rigidity. In fact, if one choose as $T$ a compact complex curve, then even $X_0=\mathbb{P}^n$ fails to work: it is well known that there exist projective bundles over compact curves that are not trivial, think of Hirzebruch surfaces.  
However, by using [Hartshorne, Deformation Theory, ex. 24.7 p. 163] one still have the following result:
Let $X_0$ be a projective scheme such that  $H^1(X_0, T_{X_0})=0$ and let $\mathfrak{X} \to T$ be a flat family of projective schemes over any nonsingular curve such that the central fibre is isomorphic to $X_0$. Then there exist a dominant base change $T' \longrightarrow T$ such that the family $\mathfrak{X}'=\mathfrak{X} \times_T T'$ is trivial over $T'$.
A: There is a notion of "deformation in the large", at least for smooth projective varieties.  A deformation in the large is any projective, $\textit{smooth}$ morphism over a connected base.  Your question does make sense for deformations in the large.  There are a number of varieties which are known to be rigid for deformations in the large, i.e., the morphism is a product étale locally over the base.  The first example is projective space (which follows, e.g., from Mori's solution of the Hartshorne conjecture).  I believe Siu first proved that hyperquadrics are rigid for deformations in the large.  And there has recently been phenomenal work of Jun-Muk Hwang and Ngaiming Mok establishing the same result for any irreducible Hermitian symmetric space, e.g., Grassmannians.
A: If $X_0$ is projectively embedded, we can use a Gr\"obner basis for its defining ideal to make a Rees family over ${\mathbb A}^1$ most of whose fibers are $X_0$, but whose central fiber is a monomial scheme. So $X_0$ must already be a monomial scheme. Since you said it's a variety, it must be a projective space. If you then allow me to reembed (by a Veronese), I could break that projective space, too, unless it's a point. So my answer: $X_0$ must be a point.
