For $k$-cycle $(1, 2, ..., k)$, do the elements in it need to be taken in order? Is $(3,1,5,2,4)$ a $5$-cycle? Or the elements in the rotation must satisfy a law of order? i.e. $5$-cycle just are $(1,2,3,4,5)$ or $(5,4,3,2,1)$.
For $k$-cycle, do the elements in them need to be in order?
When we say "$\sigma=(i_1,i_2,...,i_k)$ , then 1) $\sigma(i_1)=i_2$, $\sigma(i_2)=i_3$, ..., $\sigma(i_k)=i_1$; 2) $\sigma(n)=n$, $n\notin\left \{i_1,i_2,...,i_k\right \}$." For $\left \{i_1,i_2,...,i_k\right \}$, them can be $\left \{1, 2, 3, 4, 5\right \}$ or $\left \{1,3,2,5,4\right \}$. Is it not necessary that $i_1=1$,$i_2=2$,$\cdots$,$i_k=k$?
For the same reason that a bijection does not send a particular element in its domain to its range prior to others, the specific order in cycle notation does not matter. For example, the function (bijection) $$f: \left\{1,2,3\right\}\to \left\{1,2,3\right\}$$ given by $$1 \mapsto 3 $$ $$2 \mapsto 1$$ $$3\mapsto 2$$ is the same function $$f: \left\{1,2,3\right\}\to \left\{1,2,3\right\}$$ given by $$2 \mapsto 1$$ $$1 \mapsto 3 $$ $$3\mapsto 2$$ In other words the function (bijection) $f\in S_3$ can be represented equivalently by $$(1 3 2)$$ or $$(2 1 3).$$