G-equivariance at the Lie algebra level Let $\vec{x}, \vec{p} \in \mathbb{R}^3$ be coordinates on 6D phase space.
Consider the lift of the standard action of SO(3):
$$\psi_O(\vec{x}, \vec{p}) = (O\cdot\vec{x}, O\cdot\vec{p})
$$
This action is hamiltonian with comoment map L given by the angular momentum:
$$L(\xi) = \vec{L}\cdot\vec{\xi}
$$
where o(3) has been identified with $\mathbb{R}^3$ with Lie bracket given by the cross product.
This is a Lie Algebra homomorphism from $\mathfrak{g}$ to $C^\infty(M)$
$$\{L(\xi), L(\eta)\} = L ([\xi, \eta])
$$
The existence of the comoment map implies the existence of the moment map and hence SO(3) equivariance of $L$.
$$L\circ \psi = \text{Ad}^* \circ L
$$
And so in this context I can see
$$\text{LA homomorphism} \quad\Longleftrightarrow \quad \text{equivariance}
$$

Now, I am working through Variations on a theme by Kepler, Guillemin and Sternberg.
They give another comoment map:
$$F(\xi) = \vec{F}\cdot\vec{\xi}$$
where $\vec{F} = \vec{p} \times \vec{L}$ is the Lenz vector.
One can check the map $F$ is SO(3)-equivariant by showing the adjoint map of SO(3) on o(3) is the standard action of SO(3) on o(3) identified with $\mathbb{R}^3$.
The result then follows from the fact SO(3) preserves the cross product.
The authors give as a consequence of this:
"This map ($F$) is SO(3) equivariant, which implies, at the Lie algebra level, that:"
$$\{L(\xi), F(\eta)\} = F([\xi, \eta])
$$
Although this appears natural I am not sure how I can prove this.
I can see the difference with the previous section is that this time the map $F$ is equivariant w.r.t. the action generated by $L$ (and not by F itself), the standard action of SO(3) given at the beginning:
$$F \circ \psi^L = (\text{Ad}_L)^* \circ F
$$
It appears the result of the first section is a special case where the map generates the action under which it is equivariant.
Any tips on proving $\{L(\xi), F(\eta)\} = F([\xi, \eta])$ as a consequence of SO(3) equivariance of the map $F$ would be great.
 A: Given $\xi\in\mathfrak{so}(3)$, let $\hat{\xi}$ denote the vector field on $T^*\mathbb{R}^3$ generated by $\xi$, i.e.
$$
\hat{\xi}(x,p) := \frac{d}{dt}\Bigg\vert_{t=0}(\exp(t\xi)x, \exp(t\xi)p) = (\xi x, \xi p).
$$
The essential idea is that $L(\xi)$ is the Hamiltonian corresponding to Hamiltonian vector field $\hat{\xi}$, meaning that (with Guillemin and Sternberg's conventions}
$$
\hat{\xi} = -\{L(\xi), \cdot\}
$$
(note that the right hand side is a mapping from $C^\infty(T^*\mathbb{R}^3)$ to itself that satisfies the Leibniz property, and hence is a vector field on $T^*\mathbb{R}^3$, so this makes sense). Given this, the result you are trying to prove can be reformulated as
$$
\hat\xi(F(\eta)) = -F([\xi,\eta]) = F(-\operatorname{ad}_\xi\eta) = (-\operatorname{ad}_\xi^*F)\eta
$$
i.e.
$$
\hat{\xi}F = -\operatorname{ad}_\xi^*F.
$$
This is simply the infinitesimal version of the equivariance property
$$
F\circ \psi_O = \operatorname{Ad}_{O^{-1}}^*F
$$
(just take $O = \exp(t\xi)$, and differentiate with respect to $t$ at $t=0$).
So if you accept the latter, the former follows immediately.
