Symmetry dictates the answer I'm quite new to Electrodynamics and while I was reading about that topic (at http://textbook.scotthill.us/05-FieldLines/03-Symmetry.php) I came across the following statement:

An electric field has the same symmetries as the source charges that created it.

I can see it holds for rotation and reflection symmetries, but does it hold for every symmetry (scaling and translation, for example)?
Furthermore, is it there any proof for this?
 A: Consider the transformation: $A: \mathbb{R}^3 \mapsto \mathbb{R}^3$ that leaves the static charge density preserved 
$$(A\rho)(\vec{r}) = \rho(\vec{r}).$$
Next is the Poisson equation
$$ (\nabla^2 V)(\vec{r}) = - \frac{\rho(\vec{r})}{\epsilon_0} $$ with appropriate boundary conditions. If $\rho$ is a symmetry of $A$, then $ \nabla^2 V$ is a symmetry as well. This is shown by acting on the Poisson equation with $A$. To prove that $V$ is a symmetry one must first prove that boundary conditions remain the same. First boundary condition $\displaystyle{\lim_{r->\infty} V(\vec{r}) = const}$ certainly remains the same. 
Second Dirichlet/Neumann boundary condition (See Jackson - Classical Electrodynamics 3rd edition) is specifying potentials/potential normals along the surfaces of conductors. But if $\rho$ on conductors is unchanged than their surfaces must also be unchanged (because charge can only gather at the surface of the conductor) - hence the boundary conditions are unchanged. Considering the uniqueness of the Poisson equation one can conclude that electrostatic potential $V$ remains the same.
I only spoke in terms of conductors and densities. What about point charges? You could approximate a point charge with a small conducting sphere of small radius $\eta$ with surface potential $\frac{q}{4\pi \epsilon_0 \eta}$ and let $\eta \rightarrow 0$ which is covered by the case above.
