I am tasked to find the Fourier transform of $f(t) = e^{-3t^2}$ when it is known that $g = \hat{g}$ with $g(t) = e^{-\pi t^2}$. To be specific, beyond the definition of a Fourier transform and the invariantness of $g$, I am lead to believe that these are the only two things we known in the sense that if you'd like to use the fact that $\int_{\mathbb{R}}e^{-x^2}dx = \sqrt{\pi}$, you would have to either reason about it or prove it. And this was the way I originally solved the problem. Namely, I showed that for $a > 0$, the Fourier transform of the function $h(t) = e^{-at^2}$ is $\hat{h}(v) = \sqrt{\frac{\pi}{a}}e^{(\pi v)^2/a}$.
But I feel like this is "too much" work due to the phrasing of the problem and I would like to see some alternative way to solve it.
One thing we see immediately is that for $c = \pm \sqrt{\frac{\pi}{3}}$, $f(ct) = e^{-\pi t^2}$, whence $\hat{f(ct)}(v) = e^{-\pi v^2}$. So how could we pull out the factor of $c$ from the transform?
Indeed, there is no complicated integral to compute. It is sufficient to use the fact that for any function $f = f(t)$ in dimension $1$ and any $a\neq 0$, it follows by a simple change of variable that $$ \widehat{f(at)}(v) = \int_{\Bbb R} f(at) \,e^{-2i\pi \,t\cdot v}\,\mathrm d t = \frac{1}{|a|}\int_{\Bbb R} f(t) \,e^{-2i\pi \,t\cdot v/a}\,\mathrm d t = \frac{1}{|a|}\,\widehat{f(t)}(v/a). $$ In particular, since $e^{-3t^2} = e^{-\pi\left(at\right)^2} = g(at)$ with $a = \sqrt{3/\pi}$, it yields $$ \widehat{e^{-3t^2}}(v) = \widehat{g(at)}(v) = \frac{1}{|a|}\,\widehat{g(t)}(v/a) $$ and so since $\widehat{g} = g$, $$ \widehat{e^{-3t^2}}(v) = \frac{1}{|a|}\,g(v/a) = \sqrt{\frac{\pi}{3}}\,e^{-\pi^2\,v^2/3}. $$