= Solution
Use $\omega=\sum_i dq_i\wedge dp_i$, $\iota_{X_f}\omega=df$ and
$$
\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right).
$$
Then $X_f(g)=\{g,f\}$ and $[X_f,X_g]=-X_{\{f,g\}}$. For a left <group action>, let $\xi_P(p)=\left.\frac{d}{dt}\right|_0\exp(t\xi)\cdot p$; its fundamental fields obey $[\xi_P,\eta_P]=-[\xi,\eta]_P$.
The components of a <moment map> satisfy
$$
\mu_\xi(p)=\langle\mu(p),\xi\rangle,\qquad
d\mu_\xi=\iota_{\xi_P}\omega.
$$
Thus $\xi_P=X_{\mu_\xi}$. A <Hamiltonian> <moment map> additionally has coadjoint equivariance,
$$
\mu(gp)=\operatorname{Ad}_{g^{-1}}^*\mu(p).
$$
Here the right side evaluates on $\xi$ as $\langle\mu(p),\operatorname{Ad}_{g^{-1}}\xi\rangle$. With that convention its infinitesimal form is $\{\mu_\xi,\mu_\eta\}=\mu_{[\xi,\eta]}$. Some definitions reserve “<moment map>” for this equivariant version; choosing Hamiltonians for every generator first gives a <weakly Hamiltonian action>.
To derive the obstruction, compare the two antihomomorphism identities:
$$
X_{\{\mu_\xi,\mu_\eta\}-\mu_{[\xi,\eta]}}=0.
$$
On connected $P$, nondegeneracy of $\omega$ makes
$$
\kappa(\xi,\eta)=\{\mu_\xi,\mu_\eta\}-\mu_{[\xi,\eta]}
$$
a constant alternating bilinear form. The two Jacobi identities imply
$$
\kappa([\xi,\eta],\zeta)+\kappa([\eta,\zeta],\xi)+\kappa([\zeta,\xi],\eta)=0.
$$
Replacing $\mu_\xi$ by $\mu_\xi+b(\xi)$ changes $\kappa$ to $\kappa-b([\xi,\eta])$. Thus vanishing of this class in <Lie algebra cohomology> is the precise <moment-map equivariance obstruction>. In particular \b[a connected group with $H^2(\mathfrak g,\mathbb R)=0$ permits constants to be chosen so that the component <Poisson algebra> is the <Lie algebra>]. A locally effective action makes this realization faithful; otherwise it realizes the quotient by the infinitesimal kernel. The linear span of the components has the Lie brackets; the entire algebra of polynomial products of them is, of course, larger.
A useful sufficient condition is semisimplicity, which can be proved here without assuming the desired equivariance. Let $B$ be the nondegenerate <Killing form> and define $D$ by $B(D\xi,\eta)=\kappa(\xi,\eta)$. Invariance of $B$ and the cocycle identity give
$$
B(D[\xi,\eta],\zeta)
=B(D\xi,[\eta,\zeta])+B(D\eta,[\zeta,\xi])
=B([D\xi,\eta]+[\xi,D\eta],\zeta),
$$
so $D$ is a derivation. Choose $Z$ uniquely by $B(Z,\xi)=\operatorname{tr}(D\operatorname{ad}\xi)$. Since $[D,\operatorname{ad}\xi]=\operatorname{ad}(D\xi)$, cyclicity of trace gives
$$
\begin{aligned}
B([Z,\xi],\eta)
&=B(Z,[\xi,\eta])
=\operatorname{tr}\bigl(D[\operatorname{ad}\xi,\operatorname{ad}\eta]\bigr)\\
&=\operatorname{tr}\bigl([D,\operatorname{ad}\xi]\operatorname{ad}\eta\bigr)
=B(D\xi,\eta).
\end{aligned}
$$
Hence $D=\operatorname{ad}Z$ and $\kappa(\xi,\eta)=B(Z,[\xi,\eta])$. The shift $b(\xi)=B(Z,\xi)$ removes the obstruction. This proves <semisimple moment-map equivariance>; for connected $G$ infinitesimal equivariance integrates along its one-parameter subgroups to group equivariance. It does not claim that arbitrary symplectic actions of arbitrary groups have <moment maps>.
For the Kepler calculation, put $k=Mm$ and work on $r=|\mathbf r|>0$, since the <Hamiltonian> is singular at the origin. The <Hamilton equations> are
$$
\dot{\mathbf r}=\mathbf p,\qquad
\dot{\mathbf p}=-k\frac{\mathbf r}{r^3}.
$$
The <angular momentum> therefore obeys
$$
\dot{\mathbf L}=\mathbf p\times\mathbf p+\mathbf r\times\dot{\mathbf p}=0.
$$
For the <Runge-Lenz vector>,
$$
\dot{\mathbf K}=\dot{\mathbf p}\times\mathbf L
-k\left(\frac{\mathbf p}{r}-\frac{\mathbf r(\mathbf r\cdot\mathbf p)}{r^3}\right).
$$
The <vector triple product identity> gives $\mathbf r\times\mathbf L=\mathbf r(\mathbf r\cdot\mathbf p)-r^2\mathbf p$. Therefore
$$
\dot{\mathbf p}\times\mathbf L
=k\frac{\mathbf p}{r}-k\frac{\mathbf r(\mathbf r\cdot\mathbf p)}{r^3},
$$
and the terms cancel. Since $\dot f=\{f,H\}$, this proves
$$
\boxed{\{L_i,H\}=\{K_i,H\}=0.}
$$
The conserved quantities consequently generate transformations preserving energy. Their brackets give the <Kepler dynamical symmetry algebra>. On $H<0$, define $\mathbf A=\mathbf K/\sqrt{-2H}$ as a function on that whole open region. Because $H$ Poisson commutes with both vectors, its brackets are
$$
\{L_i,L_j\}=\epsilon_{ijk}L_k,\qquad
\{L_i,A_j\}=\epsilon_{ijk}A_k,\qquad
\{A_i,A_j\}=\epsilon_{ijk}L_k.
$$
With $\mathbf J_\pm=(\mathbf L\pm\mathbf A)/2$ one obtains two commuting $\mathfrak{so}(3)$ algebras:
$$
\{J_{\pm i},J_{\pm j}\}=\epsilon_{ijk}J_{\pm k},\qquad
\{J_{+i},J_{-j}\}=0.
$$
Thus the negative-energy algebra is \b[$\mathfrak{so}(4)$]. On $H>0$, using $\mathbf A=\mathbf K/\sqrt{2H}$ instead changes the last bracket to $-\epsilon_{ijk}L_k$, yielding \b[$\mathfrak{so}(3,1)$], with $\mathbf L$ rotations and $\mathbf A$ boosts. At zero energy the brackets of conserved functions on the characteristic orbit quotient give \b[$\mathfrak e(3)=\mathfrak{so}(3)\ltimes\mathbb R^3$], with commuting $\mathbf K$.
There are two global qualifications. First, an energy hypersurface carries a <presymplectic form>: its restricted <two-form> has characteristic direction $X_H$. At $H=0$, the literal vector fields of $\mathbf K$ need only commute modulo $X_H$, since differentiating $\{K_i,K_j\}=-2H\epsilon_{ijk}L_k$ still produces an $L_kX_H$ term. On nonzero-energy regions the energy-dependent normalization above gives an exact <Hamiltonian> <Lie algebra>; one must differentiate that normalization before restricting to a level. Second, completeness is required for a global <group action>. The collision-excluded Kepler phase space need not have complete hidden-symmetry flows, so the brackets establish local actions (or the appropriate simply connected covers), not an unconditional global action on every unregularized trajectory. Collision regularization supplies the familiar global bound-motion symmetry; this distinction is developed in https://math.berkeley.edu/~alanw/277papers00/tang.pdf . It does not change the three energy-dependent <Lie algebras> just derived.
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