Question : Let $G=\langle (123\cdots n)\rangle$ acts on $[n]$ by a natural map. Is it transitive (just one orbit)?
My Attempt : Let $G= \langle (123)\rangle $ acts on $[3]$, $G=\{id, (123),(132)\}$ so just by checking its action by brute force I come to know that it is transitive.
Second example let $G=\langle (12345)\rangle$, its action on $[5]$. Now let us see the elements of $G = \{id, (12345),(13524),(14253),(15432)\}$. Its action is also transitive.
How to prove in general that action will transitive?
Let $G$ be the cyclic group generated by the $n$-cycle $g=(12\dots n)$ for some integer $n$, and let $G$ act on $[n]$ in the natural way. I claim that for any $i,j\in[n]$ there exists some $h\in G$ such that $i^h = j$.
First, we show that it suffices to prove this is true for the case when $i=1$. Given any $i,j\in [n]$ and $h,k\in G$ such that $1^h = i$ and $1^k=j$, we have $$i^{h^{-1}k}=(i^{h^{-1}})^k = 1^k = j.$$
Thus $G$ acts transitively on $[n]$.
Now, we prove that for any $i\in[n]$ there is some $h\in G$ such that $1^h = i$ using induction. When $i=1$, we have $1^e=1$, satisfying our base case. Now, assume that for some $i\in[n]$ there is an element $h\in G$ such that $1^h= i$. Then by definition of our generator $g$, we have $$1^{hg} = (1^h)^g = i^g = i+1\!\!\!\!\pmod{n}$$
Thus $1$ can be sent to any element of $[n]$, and the result follows.