Only Legendre symbol gives constant magnitude discrete Fourier transform

Question

It is well-known that if we take the Fourier transform $\hat{g}$ of a (translated) Legendre symbol $g(x) = \pm(x+a|p)$, then $|\hat{g}(r)| = \sqrt p$ for all $r\neq 0$. I heard that the converse is also true, but this does not seems so obvious. To be precise, we want to show that for $p$ a prime and $f$ be a function on $\mathbb{Z}/p\mathbb{Z}$ that maps $0 \mapsto 0$ and $r\mapsto \pm1$ for all $r\neq 0$ the following holds. If $|\hat{f}(r)|=\sqrt p$ for all $r\neq 0$, then $f$ has the form $f(x) =\pm (x+a|p)$. Any hints on how to approach this question?