Theorem
If $p$ is odd prime then $\DeclareMathOperator{Aut}{Aut}\Aut(\bf Z_p) = (\bf Z_p)^\times =\bf Z_{p-1}$
Proof
$ G= \bf Z_p$ Let $$f_a \colon G \to G,\; f_a(x)=ax,\; 0<a<p $$
Then since $(a,p)=1$, $$ f_a(\bf Z_p) = \{0, a, 2a, \dots, (p-1)a \} = \langle a \rangle = \bf Z_p. $$ So $f_a\in \Aut (\bf Z_p)$, and $$f_a\circ f_b = f_{ab\ (p)}$$
Question
Can you finish the proof ? That is, a group $H=\{ f_a \mid 0< a < p\}$ has order $p-1$.
But I cannot show that it is cyclic
That is we must show that there exists $0< a_0 <p$ s.t. $$ \{ a_0, a_0^2, \dots, a_0^{p-1}\} = \bf Z_p^\times $$
For example, $$a_0(p=3)=2,\; a_0(5)=2,\; a_0(7)=3,\; \ldots $$
How can we show that the existence of $a_0$ ?