Regular action and integers modulo $p-1$ and $p$

Question

For $p$ a prime number, let's denote:

$$G:=\mathbb Z/(p-1)\mathbb Z = \{[i]_{p-1}\mid i=0,1,\dots,p-2\}$$

and:

$$X:=\mathbb Z/p\mathbb Z\setminus \{[0]_p\} = \{[j]_p\mid j=1,2,\dots,p-1\}$$

Does $G$ regularly (=transitively+freely) act on $X$, somehow?

My first idea was to try with $[i]_{p-1}\mapsto([j]_p\mapsto[i+j]_p)$, but of course it can't work, as $[i+j]_p=[0]_p$ as soon as $i+j=p$, and hence $[i+j]_p$ may well not even lie in $X$.

Edit. In the spirit of the question, I "don't know" that $(\mathbb Z/p\mathbb Z)^\times$ is cyclic (of order $p-1$).

Answer 1

Each group acts on itself transitively and freely, and $\Bbb Z/(p-1)\Bbb Z\cong\Bbb Z_p^*$.

Note: This is CW. Credit to @AnneBauval. See the comments.