Let $G$ be group and let $X$ be a topological space on which $G$ acts continuously. Now let us consider the following two properties relative to the group action:

(a) For every compact subset $K\subseteq X$ we have that 
$|\lbrace g\in G:gK\cap K\neq\emptyset\rbrace|<\infty$

(b) For all sequence $\{g_n\}_{n\geq 1}$ of pairwise distinct elements of $G$ and every $x\in X$ the sequence $\{g_n x\}_{n\geq 1}$ has no limit point in $X$.

It is easy to see that $(a)\Rightarrow (b)$. What about the converse? 

Under the following assumptions one may show that $(b)\Rightarrow (a)$: 

(*) Assume that $X$ is a locally compact metric space where the distance function is denoted by $d$. Assume that there is an absolute constant $C>0$ such that for all $x\in X$ there exists a neighborhood $U_x$ of $x$ such that for all $g\in G$ and all $u,v\in U_x$ one has that $d(gu,gv) < C\cdot d(u,v)$. 

For example if $G$ acts through isometries on $X$
on a locally compact space then $(a)$ is equivalent to $(b)$. 

1. Is it possible to weaken Assumption $(*)$ ?

2. Do we have examples of a contractible **bounded** open set $D\subseteq\mathbf{C}^n$ such that $Hol(D)$ (the group of biholomorphisms $f:D\rightarrow D$) acts transitively on $D$ but such that there exists no symmetry at a given point $x\in D$ (so at all points by homogeneity). By a symmetry at $x$ I mean an element $s\in G$ such that in a small neighborhood of $x$ only $x$ is fixed and $s^2=1$.  

If we drop the boundness condition then one may take the group $H\simeq\mathbf{C}^3$ (where the isomorphism is of course only as a complex manifold) of $3$ by $3$ Heisenberg matrices with coefficients in $\mathbf{C}$.