For connected groups, equivariance is equivalent to the moment-map Poisson bracket identity

Fix $\xi ,\zeta \in \mathfrak{g}$ and $p\in M$. By the moment equation for $\zeta$, the identity $X_{{\mu }^{\zeta }}=-{\zeta }_M$ and the Poisson convention, $d{\mu }_p^{\zeta }({\xi }_M(p))=-{\omega }_p({\zeta }_M(p),{\xi }_M(p))={\omega }_p({\xi }_M(p),{\zeta }_M(p))=\{{\mu }^{\xi },{\mu }^{\zeta }\}(p)\mathrm{.}$

Assume conversely that $G$ is connected and that the bracket identity holds on all of $M$. Fix $m\in M$ and define $u:G\to {\mathfrak{g}}^{\ast }$ by $u(g):=\mu (g\cdot m)-g\cdot \mu (m)$, the equivariance defect at $m$. Then $u$ is smooth and $u(e)=0$, and equivariance of $\mu$ is exactly the assertion $u\equiv 0$.

Fix $h\in G$ and ${\xi }_0,\xi \in \mathfrak{g}$, and put $\mathrm{\Theta }:=[{\xi }_0,{\mathrm{Ad}}_{h^{-1}}\xi ]$. By [F6], $h{\exp }_G(s{\xi }_0)={\exp }_G(s{\mathrm{Ad}}_h{\xi }_0)h$ and therefore the curve $s\mapsto h{\exp }_G(s{\xi }_0)\cdot m$ has velocity $-{\left({\mathrm{Ad}}_h{\xi }_0\right)}_M(h\cdot m)$ at $s=0$; hence ${\frac{d}{ds}\mid }_0{\mu }^{\xi }(h{\exp }_G(s{\xi }_0)\cdot m)={\omega }_{h\cdot m}({\xi }_M,({\mathrm{Ad}}_h{\xi }_0{)}_M)=\{{\mu }^{\xi },{\mu }^{{\mathrm{Ad}}_h{\xi }_0}\}(h\cdot m)={\mu }^{[\xi ,{\mathrm{Ad}}_h{\xi }_0]}(h\cdot m),$ using the moment equation, [F3] and the bracket identity.

Assume $\mu$ is equivariant, and fix $\xi ,\zeta ,p$. For all real $t$, equivariance and [F4] give ${\mu }^{\zeta }({\exp }_G(-t\xi )\cdot p)=⟨{\exp }_G(-t\xi )\cdot \mu (p),\zeta ⟩=⟨\mu (p),{\mathrm{Ad}}_{{\exp }_G(t\xi )}\zeta ⟩\mathrm{.}$ The $t$-derivative of the left side at $0$ is $d{\mu }_p^{\zeta }({\xi }_M(p))$, because $t\mapsto {\exp }_G(-t\xi )\cdot p$ has velocity ${\xi }_M(p)$ at $t=0$, and step 1.1 identifies it with $\{{\mu }^{\xi },{\mu }^{\zeta }\}(p)$. The $t$-derivative of the right side at $0$ is $⟨\mu (p),[\xi ,\zeta ]⟩={\mu }^{[\xi ,\zeta ]}(p)$ by [F5]. Since $\xi ,\zeta ,p$ were arbitrary, claim 1 holds.

With the same $h,{\xi }_0,\xi$ as in step 1.3, the second term of $u$ contributes ${\frac{d}{ds}\mid }_0⟨\mu (m),{\mathrm{Ad}}_{(h{\exp }_G(s{\xi }_0){)}^{-1}}\xi ⟩=⟨\mu (m),-[{\xi }_0,{\mathrm{Ad}}_{h^{-1}}\xi ]⟩=-⟨\mu (m),\mathrm{\Theta }⟩$ by [F5]. Since $[\xi ,{\mathrm{Ad}}_h{\xi }_0]={\mathrm{Ad}}_h[{\mathrm{Ad}}_{h^{-1}}\xi ,{\xi }_0]=-{\mathrm{Ad}}_h\mathrm{\Theta }$, step 1.3 can be rewritten as ${\mu }^{[\xi ,{\mathrm{Ad}}_h{\xi }_0]}(h\cdot m)=-⟨\mu (h\cdot m),{\mathrm{Ad}}_h\mathrm{\Theta }⟩$, and the identity $⟨h\cdot \mu (m),{\mathrm{Ad}}_h\mathrm{\Theta }⟩=⟨\mu (m),\mathrm{\Theta }⟩$ gives ${\frac{d}{ds}\mid }_0⟨u(h{\exp }_G(s{\xi }_0)),\xi ⟩=-⟨\mu (h\cdot m),{\mathrm{Ad}}_h\mathrm{\Theta }⟩+⟨\mu (m),\mathrm{\Theta }⟩=-⟨u(h),{\mathrm{Ad}}_h\mathrm{\Theta }⟩\mathrm{.}$

Fix ${\xi }_0\in \mathfrak{g}$ and consider $v(s):=u({\exp }_G(s{\xi }_0))$ as a curve in the finite-dimensional space ${\mathfrak{g}}^{\ast }$, with $v(0)=u(e)=0$ by step 1.2. For each fixed $\xi$, step 2.2 applied with $h={\exp }_G(s{\xi }_0)$ expresses $\frac{d}{ds}⟨v(s),\xi ⟩$ as a linear functional of $v(s)$ with smooth coefficients, so $v$ solves a linear ODE with continuous coefficients on every compact interval; by [F8] and $v(0)=0$ the curve $v$ vanishes identically. Hence $u$ vanishes on the whole exponential image ${\exp }_G(\mathfrak{g})$.

If $u(h)=0$ for some $h\in G$, then the same argument applied to the curve $s\mapsto u(h{\exp }_G(s{\xi }_0))$ shows $u(h{\exp }_G(\mathfrak{g}))=\{0\}$: the defect curve solves the same linear ODE and vanishes at $s=0$. The exponential image contains an open neighborhood $U$ of $e$ by [F7], is closed under inversion because $-\xi$ runs through $\mathfrak{g}$ with $\xi$, and the subgroup $H$ generated by $U$ is open (a union of translates of $U$) and closed (its complement is a union of cosets, each open). Since $G$ is connected, $H=G$ by [F7]. Every element of $G$ is therefore a finite product of elements of ${\exp }_G(\mathfrak{g})$, and induction over the factors using the vanishing statement proves $u\equiv 0$. Thus $\mu$ is coadjoint equivariant, which is claim 2.

Finally let $G$ be arbitrary and let $G^0$ be its identity component, a connected Lie group with Lie algebra $\mathfrak{g}$ and the same fundamental vector fields on $M$. Claims 1 and 2 applied with $G$ replaced by $G^0$ show that the bracket identity is equivalent to equivariance under $G^0$. Writing each $g\in G$ as $g=g_c{g}^0$ with $g_c$ a representative of $g{G}^0$ and $g^0\in G^0$, equivariance under all of $G$ is equivalent to equivariance under $G^0$ together with $\mu (g_c\cdot m)=g_c\cdot \mu (m)$ for all representatives $g_c$ and all $m\in M$.