Fix $x_0\in X$ where $X$ is a complete length space.
Define $$D:=\inf\ \bigg\{d_X(x_0,gx_0) \bigg|g\neq 1\in G\bigg\}$$ where $G \subset {\rm Isom}\ (X) $.
If $X$ is a Riemannian manifold, and if $G$ is not a continuous Lie group, then $D>0$. But if $X$ is not Riemannian manifold and if $X$ is a complete length space, then $D$ may be $0$ ?
Thank you in advance.
[Add]
Def : $G$ is discrete action on a Riemannian manifold $(M,d)$ if for each $x\in M$, there is open neighborhood $U$ s.t. $x\in U$ and $$g\in G,\ U\cap gU\neq \emptyset\Rightarrow g=e $$
Hence if $U$ contains $B_\epsilon (x)$, then $$ d(x,gx)>\epsilon$$ for all $g\neq 1$.
Def : $G\subset {\rm Isom}\ X$ where $X$ is a complete length space. If $x\in X$, then $G\cdot x$ is locally path connected if there is $\delta>0$ s.t. $d(x,g\cdot x) <\delta$ implies $$ d(x,ax),\ d(ax,gx) <0.51 \delta$$ for some $a \in G$.
So if $X$ is a Riemannian manifold, and $x\in X$, then $G\cdot x$ is locally path connected if $G$-action is discrete or $G$ is Lie group that is a connected Riemannian manifold.
Usually, $G$-action on complete length space $X$ can be thought through the following notion
$$(N_i=S^1\times \mathbb{R},G_i=\mathbb{Z}_i) \rightarrow (X=S^1\times \mathbb{R},G=S^1) $$
Example. $G$ is an uncountable proper subgroup of ${\mathbb R}$ and consider first the isometric action of $G$ on ${\mathbb R}$ (by translations). Next, at each point of $G\cdot 0\subset {\mathbb R}$ attach the unit vertical segment $\subset {\mathbb R}^2$ and equip the reulting "comb" $X$ with the Euclidean path-metric, which easily seen to be complete. The action of $G$ by translations on $X$ preserves this metric and does not extend to a (continuous) Lie group action; at the same time this action is free and every orbit is dense in the real line. In particular, for every $x$ in the real line, $\inf \{d(x,gx): g\ne e\}=0$.
Here is a related question which I do not know how to answer: Suppose that $X$ is a compact geodesic metric space, $G\times X\to X$ is an isometric free action of a compact totally disconnected group. Is it true that $G$ is finite? (This question is mildly related to the Hilbert-Smith conjecture.)