= Solution
Work away from $x=0$, where the <inverse-square potential> is singular. Variation of the <phase-space action> gives
$$
\boxed{\dot x=ep,\quad\dot p=\frac{2eg}{x^3},\quad\dot t=e,\quad\dot E=0,\quad E=H.}
$$
The last equation is the <constraint> imposed by the <Lagrange multiplier> $e$. The canonical <momentum> conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical <Poisson brackets> are
$$
\boxed{\{x,p\}=1,\qquad\{t,E\}=-1.}
$$
In particular,
$$
\{F,G\}=F_xG_p-F_pG_x-F_tG_E+F_EG_t.
$$
With the convention $\delta z=\alpha\{z,Q\}$, the <Noether charge> generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical <energy> is $E$; a convention that calls the <energy> the time-translation charge absorbs this minus sign into its parameter.
For the dilation charge,
$$
\dot D=eE-\frac12\left(ep^2+\frac{2eg}{x^2}\right)=e(E-H)=0
$$
on the <constraint surface>. Its infinitesimal canonical transformations are
$$
\boxed{\delta_D t=-\alpha t,\quad\delta_D x=-\frac\alpha2 x,\quad\delta_D E=\alpha E,\quad\delta_D p=\frac\alpha2p.}
$$
For constant $\alpha$, extending this by $\delta_De=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its <Poisson bracket> with the <energy> is
$$
\boxed{\{D,E\}=-E.}
$$
Put $C=E-H$. Direct variation of the <Hamiltonian> gives
$$
\delta_\beta H=p(tp-x)\beta-\frac{2g}{x^3}(-tx\beta)=(2tH-xp)\beta,
$$
so
$$
\boxed{\delta_\beta C=2tC\beta.}
$$
The supplied transformations are generated by the <special conformal charge of inverse-square mechanics>
$$
\boxed{K=t^2E-txp+\frac{x^2}{2},\qquad \delta_\beta z=\beta\{z,K\}.}
$$
Choose
$$
\boxed{\delta_\beta e=-2te\beta.}
$$
Then $\delta(eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot\beta$, gives
$$
\delta_\beta(\dot xp-\dot tE)=(t^2E-txp)\dot\beta-\beta x\dot x
=\dot\beta K-\frac{d}{d\lambda}\left(\frac\beta2x^2\right).
$$
Hence
$$
\boxed{\delta_\beta I=\int d\lambda\left[\dot\beta K-\frac{d}{d\lambda}\left(\frac\beta2x^2\right)\right].}
$$
For a constant parameter only the boundary term remains, identifying $K$ as the <Noether charge>. Independently, the <equations of motion> give
$$
\dot K=2teE-exp-t\left(ep^2+\frac{2eg}{x^2}\right)+exp=2te(E-H)=0.
$$
Using the same canonical <Poisson bracket> convention,
$$
\boxed{\{K,E\}=-2D,\qquad\{K,D\}=-K.}
$$
Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the <sl2R Lie algebra> of <conformal mechanics>. These are time translation, dilation and special conformal transformation. A useful normalization check is the <Casimir element>
$$
EK-D^2=\frac{x^2}{2}\left(E-\frac{p^2}{2}\right)=\frac g2
$$
on the <constraint surface>.
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