This problem is of a more arithmetic nature, than geometric.

For example, since every automorphism of $\mathbb{C}$ preserves $\mathbb{Q}$, we see that if $X$ can be defined over $\mathbb{Q}$, then $\Omega_X$ consists of a single isomorphism class. Similarly if $X$ can be defined over a number field, then $\Omega_X$ consists of only finitely isomorphism classes.

Thus the size of $\Omega_X$ is telling you something about the size of the smallest field of definition of $X$, not really anything about the geometry of $X$. My guess is that $\Omega_X$ consists of only countably many isomorphism classes if and only if $X$ may be defined over a field extension of $\mathbb{Q}$ of countable degree, or some statement of a similar vein, but I have not given it enough thought to be sure.

This problem is of a more arithmetic nature, than geometric.

For example, since every automorphism of $\mathbb{C}$ preserves $\mathbb{Q}$, we see that if $X$ can be defined over $\mathbb{Q}$, then $\Omega_X$ consists of a single isomorphism class. Similarly if $X$ can be defined over a number field, then $\Omega_X$ consists of only finitely isomorphism classes.

Thus the size of $\Omega_X$ is telling you something about the size of the smallest field of definition of $X$, not really anything about the geometry of $X$.

On the other hand given any two algebraically independent transcendental numbers $\alpha$ and $\beta$, there is an automorphism of $\mathbb{C}$ which swaps $\alpha$ and $\beta$, and moreover there are uncountably many such numbers. Therefore, my guess is that $\Omega_X$ consists of countably many isomorphism classes, iff $\Omega_X$ consists of finitely many isomorphism classes, iff $X$ may be defined over a number field.