Combinatorics: difference of permuted elements is constant

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This is basically an equivalent problem of another problem, and I introduce the problem in this form as I think this is more generally understandable.

Let $p$ be an odd prime, and $i, j$ be distinct elements among $S=\{0, 1, \cdots, p-1\}$, none of which are zero. Suppose there exists a permutation $\sigma : S \rightarrow S$ such that the following holds (from here all values are taken modulo $p$);

i) The value of $|\sigma(k+1)-\sigma(k)|$ is either $1$ or $j$ for all $k\in S$.

ii) The value of $|\sigma(k+i)-\sigma(k)|$ is either $1$ or $j$ for all $k\in S$.

My suggestion is that we either have two cases; that is,

i) $|\sigma(k+1)-\sigma(k)|=1$ and $|\sigma(k+i)-\sigma(k)|=j$ for all $k \in S$

ii) $|\sigma(k+1)-\sigma(k)|=j$ and $|\sigma(k+i)-\sigma(k)|=1$ for all $k \in S$

While the conclusion looks pretty intuitive, I'm stuck with how to prove it. Any helps would be greatly appreciated.