Suppose I am working with a category of objects such that each object $X$ in the category has as a trivial automorphism group. An example of such an object would be genus zero curves with $n$-punctures, i.e., $(\mathbb{P}\_k^1, (s\_i)\_{i=1}^n \in k )$. 


Let $F(S)/{\sim}$ represent families of these objects over an affine scheme $S=\operatorname{Spec} A$ where $A$ is a $k$-algebra and $k$ an algebraically closed field. As an example, consider families of genus zero curves with $n$-punctures over $S$ and fibers $X= X\_s \cong (\mathbb{P}\_k^1, (s\_i: k \to \mathbb{P}\_k^1)\_{i=1}^n) $. Then $H^1(X, TX)=0$ and $Aut(X)=0$. 


**Motivation for question**: It seems that since both $H^1(X, TX)=0$ and $Aut(X)=0$ we can conclude that every family of genus zero curves with $n$-punctures is trivial, i.e any family over $S$ is isomorphic to the trivial family $(\mathbb{P}\_A^1, (s\_i: A \to \mathbb{P}\_A^1)\_{i=1}^n)$. 


**Question**: Now going back to the beginning, suppose my family of objects over $S$ is such that all fibers are isomorphic to some objet $X$, $H^1(X, TX)=0$ and $Aut(X)=0$, does this imply that all families are trivial? 


Or even more generally, suppose $H^1(X, TX)=n$ and $Aut(X)=0$, does this imply that there exists $n$ distinct families $F(S)/{\sim}$? 


*I know there are probably many conditions on $X$ I should include, however, just assume that my object $X$ has any property that makes it "nice" to work with.*



Objects with trivial automorphism group
No. An example of a non-trivial family would occur with $S=\mathbb{A}^1\setminus\{0,1\}$ and $X\_s$ given by $\mathbb{P}^1$ marked at the points $(0,1,\infty,s)$. I think the issue you run into is basically what S. Carahan noted; if $X$ were a curve, its infinitesimal deformations would be given by $H^1(X,TX)$, but you've also bundled in the data of the points. So the first-order deformation space is instead $H^1(X,T\mathbb{P}^1\otimes \mathcal{O}(-p\_1-\ldots-p\_n))$, which is nonzero for $n\geq 4$.


I'm not quite sure what your second question is asking.


