Let $M$ be a Riemannian manifold and $G$ a subgroup of $\text{Iso}(M)$.
I want to show the following equivalence:
The quotient map $\pi:M\rightarrow M/G$ is a local isometry if and only if every point in $M$ has a neighbourhood in which each point belongs to a different orbit of $G$.
Since every point $x\in M$ has a neighborhood $U_x$ such that $y\in U_x$ implies that $g\in G, g\neq Id g(y)$ is not in $U_x$, this implies that $g(U_x)\cap U_x=\phi$ if $g\neq Id$. We can assume that $U_x$ is the domain of the chart $(U_x,f_x)$. Let $p:M\rightarrow M/G$ the quotient map. There exists a structure of a differentiable manifoldon $M/G$ whose atlas is defined by $(\pi(U_x),f_x\circ \pi^{-1}_{\mid \pi(U_x)})$. Where $\pi^{-1}_{\pi\mid \pi(U_x)}:\pi(U_x)\rightarrow U_x$ is the inverse of $p$.
There exists a metric of $M/G$ defined by $\langle u,v,\rangle_{\pi(x)}=\langle u',v'\rangle_x$ where $dp_x(u')=u$, $dp_x(v')=v$, this definition does not depend of the point in the fiber of $\pi(x)$ since $G$ acts by isometry and $\pi$ is an isometry if $M/G$ is endowed with this metric.