I have just learned about the Fourier transform, and a I have thought of a nice question i cannot answer. The Fourier transform of a Gaussian curve, $e^{-{\alpha}x^2}$ is $\frac{1}{\sqrt{2\alpha}}e^{-\frac{\omega^2}{4\alpha}}$ which is essentially the same "shape". Another example which we saw is a chain of delta functions, $\sum_{n=-\infty}^{\infty}{\delta(x-nA)}$ (for some $A\in\mathbb{R}$), which its Fourier transform is $\frac{\sqrt{2\pi}}{A}\sum_{n=-\infty}^{\infty}{\delta(\omega-\frac{2\pi}{A}n)}$. Again, the same "shape". What's I'm asking is are there any other known functions which theirs Fourier transform has the same shape?
By "shape" I mean that if $f(x)=af(bx+\phi)$ then the Fourier transform of $f$ will be $\hat{a}f(\hat{b}\omega+\hat{\phi})$ for some $\hat{a}, \hat{b}, \hat{\phi}$.
TL DR:
I will show that the only integrable functions that satisfy the relation are the eigenvectors of the Fourier transform (either the continuos or the discrete one) that are characterized here.
As the Fourier transform is linear we can assume $a =1$. Let $b\ne 0$ and assume $$ \mathcal{F}[f(bx+\phi)]=a'f(b'\xi+\phi') $$ By the involutive property of the Fourier transform $$ \mathcal{F}^2f(bx+\phi)=f(-bx-\phi)=\mathcal{F} \left[a'f(b'\xi+\phi')\right]=\mathcal{F}\left[a'f\left(b\left(\frac{b'\xi+\phi'-\phi}{b}\right)+\phi\right)\right] $$ so $$ f(-bx-\phi)=a'^2f\left(\frac{b'^2}{b} x +\left(\frac{b'}{b}+1\right) \phi' -\frac{b'}{b} \phi\right) $$ or $$ f(x)=a'^2f\left(-\left(\frac{b'}{b}\right)^2 x -\frac{1}{b}\left(\frac{b'}{b}+1\right) \phi' +\frac{1}{b}\left(\frac{b'}{b}-1\right) \phi\right) $$
If $\frac{b'}{b} \ne \pm 1$ it is easy to see that the function is continuos iif it is constant. As a $L^1(\mathbb{R}^n)$ function has a uniformly continuous Fourier transform, this implies that there are no $L^1$ functions that satisfy the above inequalities.
Suppose $\frac{b'}{b}=1$. the equation becames: $$ f(x)=a'^2 f\left(x+\frac{2}{b} \phi'\right) $$
If $\phi' \ne 0$ then, suppose $a'^2 \ne \pm 1$ and consider the sequence $x_0=x$, $x_n= x +n\frac{2}{b} \phi'$. We have $\lim_{n \to \infty} x_n = \pm \infty$ (depending from the sign of $\frac{\phi'}{b}$ and $$ f(x_n)= a'^{2n} f(x_0) $$ By the continuity of $f$ it follows that it growth exponentially at infinity i.e. it isn't a tempered distribution, but this is impossible as $f \in L^1$. So $(a')^2= \pm 1$ that implies that $f$ is periodic. In particular, it is an eigenvector of the discrete Fourier transform (Indeed $a'= \pm 1$ or $\pm i$).
If $\phi'=0$. As above we have $$ (\mathcal{F}^{-1})^2 f(b' \xi) = f (-b' \xi) = \cdots =a'^2 f\left(bx+ 2 \phi \right) $$ From this you obtain that if $\phi \ne 0$ you have $a'= \pm 1$ or $\pm i$ and $f$ is an eigenvector of the discrete Fourier transform.
If $\phi=\phi'=0$ you obtain $$ \mathcal{F}f(bx)=a'f(bx) $$ So $f$ is an eigenvector of the continuos Fourier transform.
If $\frac{b}{b'}=-1$, by interchanging the roles of $\phi$ and $\phi'$ in the previous proof we see that if at least one of the two phases is not zero, than $f$ is an eigenvector of the discrete Fourier transform.
If $\phi=\phi'=0$ then
$$ \mathcal{F} f(bx)= a'f(-bx) $$ By the involutive property of the Fourier transform it follow that $f$ is even and so it is an eigenvector of the continuous Fourier transform