Proving that a generic variety with ample canonical bundle has no automorphisms Let $X$ be a smooth projective connected variety over the complex numbers with ample canonical bundle. If $X$ is generic and $\dim X \leq1$, the automorphism group of $X$ is trivial, see for instance
Why is a general curve automorphism-free?
This question is about generalizing this to arbitrary dimension. Let me be more precise.
Suppose that $X$ is "generic". Is the automorphism group of $X$ trivial?
This is probably true, and there are three approaches to this sketched in the above MO question. The first two might not be feasible.


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*Use deformation theory, i.e., compute the tangent space at the moduli space, and use Lefschetz trace formula. Can somebody make this more precise in this case?

*Count parameters using Riemann-Hurwitz. This is going to be problematic in the higher-dimensional case, even though there is a Riemann-Hurwitz formula, I am not sure the dimension of the moduli space is explicitly known (as opposed to the one-dimensional case where it equals $3g-3$).

*Exhibit an $X$ as above with trivial automorphism group for any possible hilbert polynomial. In fact, the order of the automorphism group of $X$ is bounded (even explicitly) by a constant depending only on the Hilbert polynomial of $X$.
I think 3 is the most promising, but this would require me to come up with the following.
Let h be the hilbert polynomial of $X$. Then there exists a smooth projective connected variety $Y$ with ample canonical bundle and hilbert polynomial of the canonical bundle equal to $h$ such that Aut$(Y)$ is trivial.
So my problem is to do this for every occuring hilbert polynomial. Of course, writing down varieties $X$ as above with no automorphisms is not so difficult.
 A: It seems to me that this is not true and that a counterexample can be constructed as follows.
Take a double cover $\alpha \colon X \longrightarrow A$ of an abelian surface $A$, branched over a smooth divisor $B \in |2 L|$, with $L$ very ample. We have $$K_X=\alpha^* L, \quad \alpha_* \mathcal{\omega}_X = \mathcal{\omega}_A \oplus \omega_A (L),$$ hence $$K_X^2 = 2L^2, \quad p_g(X) = 1+ h^0(A, L), \quad q(X)=2.$$ 
Than $X$ is a smooth surface of general type. Moreover, since $\alpha$ is a finite map, $X$ does not contract any curve, in particular $K_X$ is ample.
We have $q(X)=2$, so $\textrm{Alb}(X)$ is an abelian surface and, by the universal property of the Albanese map, the morphism $\alpha$ factors through $a \colon X \longrightarrow \textrm{Alb}(X)$. 
But then, since $\deg \alpha =2$ and $X$ is not an abelian surface, it follows that the isogeny $\textrm{Alb}(X) \to A$ must be an isomorphism. Then the morphism $\alpha \colon X \longrightarrow A$ coincides the Albanese map of $X$. 
By a result of Catanese (see A superficial working guide to deformation ands moduli, arXiv:1106.1368, Section 5) the degree of the Albanese map is a topological invariant.
It follows that any deformation of $X$ is still a double cover of its Albanese variety. 
Thus  any surface $Y$ lying in the same connected component of the moduli space containing $X$ has non-trivial automorphism group, because the Albanese double cover induces a non-trivial involuton $\iota \colon Y \to Y$, and so $\mathbf{Z} /2 \mathbf{Z} \subset \textrm{Aut}(Y)$.
A: As Francesco points out, the claim is  false when $\dim X>1$. The question is  discussed in 
[B. Fantechi, R. Pardini, Automorphisms and moduli spaces of varieties with ample canonical class via deformations of abelian covers, Comm.  Algebra  25 (1997), 1413-1441.  math.AG/9410006] (see in particular Thm. 6.6).
