So in the first part of the question I had to prove that $gcd(d,n)=1 \implies S_n = \langle\{(i\;\; \overline {i+d}) : 1 \le i \le n\}\rangle$ whilst $\forall a \in \mathbb {Z}$, we mark $\overline {a}$ as the only number in $\{1,2...n\}$ that satisfies $a \equiv \overline {a}\;(mod n)$.
Now for the second part which is the question I have posted I began to lose ground.
We have seen in class the following relation as to the constructor groups of $S_n$:
$S_n = \langle\{(i\;j) |\;1\le i \lt j \le n\}\rangle=\langle\{(i\;i+1)|\;1\le i \le n-1\}\rangle=\langle\{(1\;i) |\;2\le i \le n\}\rangle=\langle(1\;2),(1\;2\;3...n)\rangle$
I'm very much lost on the conncection between gcd(d,n) and the constructor group and would love some help!