On the number of $\sigma\in S_{p-1}$ of a given form.

Question

Let $p$ be a prime, $d$ a proper divisor of $p-1$, and $\sigma\in S_{p-1}$ of the form (everything is modulo $p$): $$\sigma=(1,x,\dots,x^{d-1})(i_2,i_2x,\dots,i_2x^{d-1})\dots(i_k,i_kx,\dots,i_kx^{d-1})\tag1$$ (hence $|\sigma|=d$) where $kd=p-1$ and, $\forall i,j=1,\dots,p-1$ such that $i+j\not\equiv_p0$: $$\sigma(i+j\bmod p)\equiv_p\sigma(i)+\sigma(j)\tag2$$ Does $(2)$ follow from $(1)$, or does it really establish a constraint on the possible $\sigma$'s of the form $(1)$? As for the context, I'm interested in the number $N_p(d)$ of the permutations of the form $(1)$, which fulfil $(2)$.

Answer 1

(2) does imply (1). The permutation $\sigma$ in (1) is much more conveniently written as multiplication by $x$, where $x$ has order $d$, modulo $p$. Then it's obvious that $\sigma(i+j) \equiv x(i+j) = xi+xj \equiv \sigma(i)+\sigma(j)$ (mod $p$).

Since the multiplicative group modulo $p$ is cyclic of order $p-1$, it's a standard result that the number of elements of order $d$ is exactly $\phi(d)$ (the Euler phi function); so that's how many permutations of the type (1) there are. (By the way, all of this works for $d=1$ and $d=p-1$ as well.)