It is well known that if $u$ is a harmonic function on $\mathbb R^2$ then its Kelvin transform defined by$$ v(r,\theta) = u(\frac{1}{r},\theta)$$is also harmonic for $r>0$. Note that the Kelvin transform is defined in all dimensions but I stated it in dimension two for simplicity.
My question is whether something similar can be done for the heat equation$$ \partial_t u - \Delta u =0,$$that is to say, find a transformation $\psi: \mathbb R^{1+n} \to \mathbb R^{1+n}$ so that if $u$ solves heat equation then so does $u\circ \psi$. Of course, I don't want the trivial transformations like the parabolic scaling $\psi(t,x) = (\alpha^2 t, \alpha x)$.