Following @saz answer when the transform is:
$$\hat{f}(\omega) := \int_{\mathbb{R}} f(x) \cdot e^{-\imath 2 \pi \, \omega x} \, dx$$
First we consider the case $m=0$ and $n=1$, i.e. $f(x) := \exp(-x^2)$ and $$\hat{f}(\omega) := \int_{\mathbb{R}} f(x) \cdot e^{-\imath 2 \pi \, \omega x} \, dx = \int_{\mathbb{R}} \exp \left(-x^2 \right) \cdot e^{-\imath 2 \pi \, \omega \cdot x} \, dx.$$
Differentiating with respect to $\omega$ yields $$\frac{d}{d\omega} \hat{f}(\omega) = \int_{\mathbb{R}} e^{-x^2} \cdot (-\imath 2 \pi \, x) \cdot e^{-\imath 2 \pi \, \omega x} \, dx = \imath \pi \int_{\mathbb{R}} \left( \frac{d}{dx} e^{-x^2} \right) \cdot e^{-\imath 2 \pi \, \omega x} \, dx.$$
Applying the integration by parts formula, we obtain
$$\frac{d}{d \omega} \hat{f}( \omega) = -2 \pi^{2} \omega \cdot \int_{\mathbb{R}} e^{-x^2} \cdot e^{-\imath 2 \pi \, \omega \, x} \, dx =-2 \pi^{2} \omega \cdot \hat{f}( \omega).$$
The unique solution to this ordinary differential equation is given by
$$\hat{f}(\omega) =c \cdot \exp \left( -\pi^{2} \omega^{2} \right).$$
Since $c=\hat{f}(0) = \int_{\mathbb{R}} f(x) \, dx$, it follows that $c = \sqrt{\pi}$. Moreover, applying the following well-known formulas
$$\begin{align} \widehat{f(x+m)}(k) &= e^{\imath \, k \cdot m} \hat{f}(k) \\
\widehat{f(\alpha \cdot x)}(k) &= \frac{1}{\alpha} \cdot \hat{f} \left( \frac{k}{\alpha} \right) \qquad \alpha>0, \end{align}$$
For $f(x)=e^{-mx^{2}}$ then
$$ \hat{f}(\omega) = \frac{\sqrt{\pi}}{\sqrt{m}} \cdot \exp \left( -\pi^{2} \left(\frac{\omega}{\sqrt{m}}\right)^{2} \right) $$