The Gaussian $e^{-\pi\abs{x}^2}$ is its own Fourier transform, and its dilation $g_t$ has transform $g(t\xi)$.Suggest a slogan
Let \(g\colon\RR^d\to\RR\) be the Gaussian \(g(x) \coloneqq e^{-\pi \abs{x}^2}\). Then its Fourier transform is \(\widehat g = g\). Moreover, for \(t>0\) the dilation \(g_t(x)=t^{-d}g(t^{-1}x)\) satisfies \[ \widehat{g_t}(\xi) = g(t\xi) = e^{-\pi t^2 \abs{\xi}^2}, \qquad \xi\in\RR^d . \]