Fourier shift theorem for cyclic permutation

Question

Let $h : \Bbb R^{d} \to \Bbb R$. I am quite confused how to apply the Fourier shift theorem to $h(Px)$ where $P\in M^{d\times d}$ is a cyclic permutation acting on $x$.

$$\mathcal{F}(h(Px))(k)=?$$

I guess some phase factors should pop out, but not sure how.

Answer 1

In general for an invertible linear map $P$ the Fourier transform obeys: $$ \widehat { h \circ P} = |\det P|^{-1} \hat h \circ (P^T)^{-1} .$$ When $P$ is a permutation matrix (not just a cyclic permutation) $\det P$ is always $\pm 1$ and $P^T = P^{-1}$, so this just reduces to $$\widehat { (h \circ P )}(\xi) = \hat h (P \xi).$$