$\operatorname{Aut}_{\text{Grp}}(C_p) $ is isomorphic to $C_{p-1}$ 
Prove that if $p$ is prime, then $\operatorname{Aut}_{\text{Grp}}(C_p)$ is isomorphic to $C_{p-1}$.

Before, this proof you are asked to compute the group of automorphism of $(\mathbb{Z}, +)$. The answer I got was the identity map and $\beta : n \rightarrow -n$.
Can someone explain to me what $C_{p-1}$ is?
$\textbf{ANSWER:}$ Cp is the cyclic group of order p.
What is meant by $\operatorname{Aut}_{\text{Grp}}(C_p)$?
$\textbf{ANSWER:}$ $\operatorname{Aut}_{\text{Grp}}(C_p)$ is a group, for all $C_{p}$ of all groups.
And perhaps a sketch of the proof?
 A: I know it's been a long time. In case someone is still having troubles with excerice II.4.15 from "Algebra: Chapter 0" by P. Aluffi.
Since $p$ is prime every non-zero element of $C_p$ has order $p$(since order of elements divide order of group). Any automorphism $C_p \rightarrow C_p$ can be determined by where $g^1$ is mapped. It is easy to see that there can be $p-1$ such morphisms with identity morphism being an idendity element. This defines a group of order $p-1$.
One may treat such a group as a multiplicative group modulo $p$ (if it's not clear to you, let $\phi: G\rightarrow G \in Aut_{Grp}(G)$, with generator $g^1$ being mapped to $g^m$, then $\phi^2 = \phi \circ\phi (g) = g^{m^2}$, so it's just multiplication $mm$).
As it was pointed out in comments, in order to conclude $Aut_{Grp}(C_p)$ is cyclic of order $p-1$ we need to show that $(\mathbb{Z}/p\mathbb{Z})^*$ has an element of order  $p-1$(making it cyclic). 
Let $G = (\mathbb{Z}/p\mathbb{Z})^*, g \in G$ be an element of maximal order. Since the group is abelian the order of every element divides $|g|$. Therefore, for every $h \in G, h^{|g|} = 1$. As we can see the equation $x^{|g|} = 1$ has $p-1$ solutions modulo $p$ (because $p-1$ is group's order). From the theorem that $x^r = 1$ has at most $r$ solutions modulo $p$ we may conclude that $|g|$ is at least $p-1$, it is also at most $p-1$ as the order of group's element is at most the order of group.  
Now, suppose the order of $[m]_p$ in $(\mathbb{Z}/p\mathbb{Z})^*$ equals to $p-1$.
Then with $\phi$ defined as above $\phi^{p-1} = x^{m^{p-1}} = id$.
  Hence, you may conclude that having an element of order $p-1$ makes $Aut_{Grp}(C_p)$ cyclic of order $p-1$ and thus isomorphic to $C_{p-1}$
