Connection Between Fundamental Solution and Symmetries of PDE The typical derivation of the fundamental solution of Laplace's equation is to look for a radially symmetric solution because the Laplace equation has radial symmetry, and a similar heuristic can be used to derive the fundamental solution of the heat equation. On page $72$ of Terrence Tao's book (http://www.math.ucla.edu/~tao/preprints/chapter.pdf) he writes that one can use symmmetry methods to derive the fundamental solution of the heat equation as well,  and I have also seen similar claims elsewhere. But is this more than a heuristic? Sure we can derive the fundamental solutions using these methods, but is there any a priori reasons to believe that the solutions found should be fundamental solutions, or that fundamental solutions should possess all of the symmetries of the equation?
 A: 
Are there a priori reasons to believe that the solutions found should be fundamental solutions?

Yes. The a priori reason is that they are the fundamental solutions for the symmetry-reduced equations (usually ODEs). Of course one still needs to check that the solutions are in fact the correct ones, but once the explicit form of the fundamental solution is available the checking part is often routine. 

fundamental solutions should possess all of the symmetries of the equation?

Obviously not. The Laplace equation on $\mathbb{R}^n$ is translation symmetric. The fundamental solution $\propto r^{2-n}$ is not. One has to be a bit more careful about what you mean by symmetry. 
Suppose $L$ is a linear operator on smooth functions (perhaps with compact support) defined over some domain $\Omega$. And let $\Phi:\Omega\to\Omega$ be a symmetry of $L$, in the sense that for any smooth function $f$, $(Lf)\circ\Phi = L(f\circ\Phi)$. 
Further suppose that $L$ is injective, in the sense that $Lf = 0$ on the whole of $\Omega$ implies that $f = 0$. 
Suppose you are looking to solve 
$$ Lf = g $$
and you know that $g\circ \Phi = g$. Then by the previous discussion you know that if $f$ is a solution, then so is $f\circ\Phi$. But we then have 
$$ L(f - f\circ\Phi) = 0 \implies f = f\circ\Phi $$
The same argument essentially goes through for distributions instead of $C^\infty_c$, and in the case of the fundamental solution you take $g$ to be the $\delta$-distribution. On $\mathbb{R}^n$, $\delta$ is spherically symmetric, but not translation symmetric. Hence the fundamental solution should also be spherically symmetric, and not translation symmetric. 
