Is the inverse Fourier transform of a radial function radial?

Question

Let $f: \Bbb{R}^d \to \Bbb{R}$ be a rapidly decreasing function in $\Bbb{R}^d$. Concerning Fourier transforms in $\Bbb{R}^d$, we define $\hat{f}(\xi) = \int_{\Bbb{R}^d} f(x) e^{-2 \pi i \xi \cdot x}dx $, for $\xi \in \Bbb{R}^d$. And the inverse of $g(\xi)$ is $\int_{\Bbb{R}^d}g(\xi)e^{2 \pi i \xi \cdot x}d\xi$. A function $f: \Bbb{R}^d \to \Bbb{R}$ is said to be radial iff there is $F: \Bbb{R} \to \Bbb{R}$ such that $f(x) = F(|x|)$ for $x \in \Bbb{R}^d$. It is easy to prove that the transform of a radial function is radial. Does the same hold for inverse transforms?