Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$ and let $G=Aut(\Omega)$ be the **full** group of self-biholomorphisms of $\Omega$. Assume that there is $z\in \Omega$, such that the orbit of $z$ is somewhere dense, i. e. $\mathrm{int}~ \overline{G\cdot z}\ne\varnothing$. Does it follow that $\Omega$ is homogeneous, i.e. $G$ acts transitively on $\Omega$?



Criterion for homogeneity
**Edit: (21 May 2017)** I have modified my answer to cover the case that the OP meant to ask, i.e., the assumption is that the *closure* of an orbit has nonempty interior.


Now that you have added the assumption of boundedness (and, I assume, connectedness, since, without it, the answer would clearly be 'no'), we can answer the question: Yes, a connected, bounded domain $\Omega\subset \mathbb{C}^n$ is homogeneous as long as there is any point $z\in\Omega$ such that $\mathrm{int}\bigl(\,\overline{G{\cdot}z}\,\bigr)$ is not empty.


The argument is as follows: By a result of Kobayashi $\Omega$ is hyperbolic, and hence $G=\mathrm{Aut}(\Omega)$ is a Lie group that, in particular, preserves the Bergman metric on $\Omega$ (cf., Theorem 9.1 of S. Kobayashi, *Intrinsic distances, measures, and geometric function theory* Bulletin of the AMS **82** (1976), 357–416). 


Suppose that $z\in \Omega$ is such that the closure of its orbit contains an $\epsilon$-ball $B\_\epsilon(p)\subset \Omega$ for some positive $\epsilon$ (using, say, the Bergman metric to define distances). Then the closure of this orbit contains $B\_{\epsilon/2}(g{\cdot}z)$ whenever $g{\cdot}z \in B\_{\epsilon/2}(p)$, and hence the closure of the orbit of $z$ contains $B\_{\epsilon/2}(g{\cdot}z)$ for all $g\in G$. Consequently, the closure of the orbit of $z$ is both open and closed and hence, by connectedness, is all of $\Omega$.


Now, let $p\in\Omega$ be arbitrary and let $g\_k\in G$ be a sequence such that $g\_k{\cdot}z$ is a Cauchy sequence that converges to $p$. Since the maps $\phi\_k:\Omega\to\Omega$ defined by $\phi\_k(w) = g\_k{\cdot}w$ are isometries of the Bergman metric, it follows that we can, by reducing to a subsequence, assume that,
not only does $\phi\_k(z)$ converge to $p$, but the sequence $\phi\_k'(z):T\_z\Omega\to T\_{\phi\_k(z)}\Omega$ converges to an isometry $f:T\_z\Omega\to T\_p\Omega$. It then follows by a standard argument using the path-connectedness of $\Omega$, that the sequence of Kähler isometries $\phi\_k:\Omega\to\Omega$ converges on compact sets to a Kähler isometry $\phi\_\infty:\Omega\to\Omega$. 


It is easy to show that $\phi\_\infty$ is a bijection, and, hence, it must belong to $G$. Thus, $p$ is in the $G$-orbit of $z$. Since $p$ was arbitrary, it follows that $G{\cdot}z = \Omega$, as was to be shown.


