electromagnetism - Special conformal transformations and image charges

Let us consider a grounded conducting sphere with radius $r$ and a point charge $e$ at a distance $R>r$ from the center of the sphere. For simplicity, we can choose the sphere to be centered at the origin and the charge to be on the $z$ axis at $(0,0,-R)$.

This problem can be solved by the method of image charges, i.e. by replacing the sphere with another point charge $q=-er/R$ at $(0,0,-r^2/R)$. Indeed, the resulting electrostatic potential $$\Phi(x,y,z)=\frac{e}{4\pi}[x^2+y^2+(z+R)^2]^{1/2}-\frac{er}{4\pi R}[x^2+y^2+(z+r^2/R)^2]^{1/2}$$ solves the Poisson equation $$\Delta \Phi =- e\, \delta(x,y,z+R)$$ with boundary condition $$\Phi=0\qquad \text{at }|\mathbf x|=r\,.$$ Furthermore, the charge surface density on the sphere is given by $$-\left(\frac{\partial\Phi}{\partial \rho}\right)_{\rho=r}=-\frac{er}{4\pi}\frac{(R/r)^2-1}{(r^2+2rR\cos\theta+R^2)^{3/2}}$$ in spherical coordinates $\rho,\theta,\phi$.

We might also tackle this problem by exploiting the conformal symmetry electromagnetism: the action $$\int d^4x \sqrt{-g} F_{\mu\nu} F^{\mu\nu}-\int d^4x \sqrt{-g} A_\mu j^\mu$$ is invariant under local Weyl rescalings $g_{\mu\nu}\mapsto \Omega^2 g_{\mu\nu}$ if $$A_{\mu}\mapsto A_{\mu}\,,\qquad j_{\mu}\mapsto \Omega^{-4} j^\mu\,.$$ Performing a special conformal transformation with parameter $b^\mu=(0,0,0,1/r)$, i.e. $$t'=\frac{t}{\sigma(t,\mathbf x)}\,,\qquad x'=\frac{x}{\sigma(t,\mathbf x)}\,,\qquad y'=\frac{y}{\sigma(t,\mathbf x)}\,,\qquad z'=\frac{z-\mathbf (x^2-t^2)/r}{\sigma(t,\mathbf x)}$$ with $$\sigma(0,\mathbf x)=1-2z/r+(\mathbf x^2-t^2)/r^2$$ we see that the sphere in the un-primed space corresponds to the plane $$z'=-\frac{r}{2}$$ in the primed space. The corresponding conformal factor can be read off the transformation rule $$g'^{00}=\left(\frac{\partial t'}{\partial t}\right)^2 g^{00}=\frac{g^{00}}{\sigma^2(0,\mathbf x)}\qquad \text{at }t=0.$$ Hence, $g'_{\mu\nu}=\sigma^{2}g_{\mu\nu}$ and the Weyl rescaling needed to go back to the original Minkowski metric $g_{\mu\nu}$ is $g'_{\mu\nu}\mapsto \sigma^{-2}g'_{\mu\nu}$. This gives $\Omega=1/\sigma$.

Now I would like to find the rule for transforming the charge $e$ and its image $q=-er^2/R$. Assuming that a point-like charge $e$ at $\mathbf x_0$ scale as $$\frac{e}{\sigma(0,\mathbf x_0)^{1/2}}$$ under conformal transformations, $$e\longmapsto e'=\frac{e}{\sigma(0,0,0,-R)^{1/2}}=\frac{e}{1+R/r}$$ and we get a charge $e'$ at $z'=-rR/(r+R)$; then, the image $q'$ of $e'$ with respect to the plane is $$q'=-e'=-\frac{e}{1+R/r}$$ at $z'=-r^2/(R+r)$. And the indeed the matching is correct, because the conformal mapping of $q$ according to the above rule is precisely $$q\longmapsto q'=\frac{q}{\sigma(0,0,0,-r^2/R)^{1/2}}=-\frac{e}{1+R/r}$$ at $z'=-r^2/(R+r)$.

My question is, why is the above transformation rule for $e$ hold? Naively I would have expected $$j^0(x) = e \delta(\mathbf x -\mathbf x_0)\longmapsto \Omega^{-4} \frac{\partial t'}{\partial t} e \delta(\mathbf x(x')-\mathbf x_0)=\sigma^4 \sigma^{-1} \sigma^{-3} e \delta (\mathbf x'-\mathbf x_0')$$ but this would mean that they do not transform...