In Berndt's $\textit{Introduction to Symplectic Geometry}$ we have the following statements:
Assume we are given a symplectic manifold $(M, \omega)$, a symplectic operation $\phi:G\times M \rightarrow M$ of the Lie group $G$ on $M$ and for $\mathfrak{g}$ (the Lie algebra of $G$) an associated $\textrm{Ad}^*$-equivariant moment map $\Phi:M\rightarrow \mathfrak{g}^*$. Then we denote by $G_\mu$ for $\mu \in \mathfrak{g}^*$ the isotropy group $$G_\mu:=\{g\in G; \textit{ Ad}_{g^{-1}}^*\mu=\mu\}.$$ Since $\Phi$ is $\textrm{Ad^*}$ equivariant, the space $M_\mu:=\Phi^{-1}(\mu)/G_\mu$ of $G_\mu$ orbits on the fibers $\Phi^{-1}(\mu)$ makes sense.
$\textbf{Question(s):}$ What do the elements of $G_\mu$ and $M_\mu$ look like? How does the equivariance condition help here?
Equivariancy of $\Phi$ implies that for $g\in G_\mu$ and $x\in M$ with $\Phi(x)=\mu$, we get $\Phi(\phi(g,x))=Ad_{g^{-1}}^*(\mu)=\mu$. Thus the level set $\Phi^{-1}(\mu)$ is invariant under the action of the subgroup $G_\mu\subset G$, so it decomposes into $G_\mu$-orbits and it makes sense to look at the space of orbits.