Can one understand the Kelvin transform conceptually?

Let $U = \mathbf{R}^n - \{ 0 \}$, $n > 2$ and consider for a function $f \in C^2(U)$ the Kelvin transform

$$f^\star(x) = r^{2-n} f\left(\frac{x}{r^2}\right),$$

where $r = \lvert x \rvert$. One can verify by explicit computation that,

$$r^2 \Delta(f^\star(x))= (r^2\Delta f)^{\star}(x),$$

where $\Delta$ denotes the usual Laplace operator. In particular the Kelvin transform maps harmonic functions to harmonic functions. Is there a way to see this last fact without resorting to explicit computation? I suspect it has to do with how the group of conformal transformations acts.

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Yes, that's it. You can think of $\mathbb{R}^n \setminus \{0\}$ as being stereographically projected from the sphere. Then the Kelvin transform is just "flipping" the sphere. –  Ray Yang Jun 11 '12 at 7:35