I played a little with convolutions of the form $$g(\xi) = \int_{-\infty}^\infty f(x) f(\xi - x) \mathrm{d}x,$$ where $f(x)$ should be a rapidly decaying function (not necessarily in a Schwartz sense, it should be enough that $f(x)$ is integrable).
Concretely, I looked at these examples
\begin{align*} f_1(x) &= e^{-|x|} & g_1(\xi) &= (1+|\xi|)e^{-|\xi|} \\ f_2(x) &= e^{-x^2} & g_2(\xi) &= \sqrt{\frac\pi 2}e^{-\xi^2/2} \\ f_3(x) &= \frac{1}{x^2+1} & g_3(\xi) &= \frac\pi 2 \frac{1}{1+(\xi/2)^2} \\ f_3(x) &= \frac{1}{x^4+1} & g_3(\xi) &= \frac{\pi}{8\sqrt{2}} \frac{12+\xi^2}{2+\xi^2} \frac{1}{1+(\xi/2)^4} \\ f_4(x) &= Ai(x) & g_4(\xi) &= Ai(\xi/\sqrt[3]{2}), \end{align*} where $Ai(x)$ is the Airy function.
Noticeably, it seems in every case that $g(\xi) = h(\xi) f(\alpha\xi)$, where $\alpha \in \mathbb{R}$ and $h(\xi)$ is some function. This is a statement, which I very much like to proof, or, at least, give some conditions, for when this holds. However, I wasn't very successful so far.
Obviously, for rational functions like $f_3$ and $f_4$ the convolution theorem provides an easy answer, yet this seems to be true for even a wider class of functions.
Side Note Actually, this topic sounds so simple that I assume that this must have been asked somewhere already, yet I did not manage to find anything on this problem
Obviously you could always simply choose $h(\xi) = g(\xi)/f(\alpha\xi)$, so the question is in its current form not well posed.
Moreover note that for example for $f(x) = |x|e^{-|x|}$ one has $g(\xi) =\tfrac 16 (|\xi|^3 + 3|\xi| + 3)e^{-|\xi|}$, which does not exhibit the pattern you observed without resorting to the 'trivial' choice of $h$ pointed out above.