If a finite group $G$ has a normal subgroup $N$ of order $3$ that is not contained in $Z(G)$, then $G$ has a subgroup of index $2$. 
If a finite group $G$ has a normal subgroup $N$ of order $3$ that is not contained in $Z(G)$, then $G$ has a subgroup of index $2$.
Since $N$ is normal, we can consider the action of $G$ on $N$ by conjugation, because this will induce a homomorphism that is present in any such mapping of a group into its normal subgroups.

I understand the concept of this, but I am looking at one proof that defines the homomorphism to map from $G$ to $\operatorname{Aut}(N)$. Why is the codomain of this mapping $\operatorname{Aut}(N)$ instead of N?
Then the proof will claim that $\ker \phi = C_{G}(N)$. How do we know that this is true?
The principles used thereafter are just counting, so I can understand it from there. We will say that since $|N|=3$ it contains exactly the elements $1,n,n^{-1}$. Then with the identity it is obvious that for all $g \in G$, we know $g \cdot 1 = g1g^{-1}=1$, and so either $g \cdot n = n$ or $g \cdot n = n^{-1}$. Thus, $\lvert\operatorname{Aut}(N)\rvert=2$, so $G/C_{G}(N)=2$. Therefore the centralizer is a subgroup of order $2$.
 A: Let me give some structure to your problem:
You have 


*

*a finite group $G$.

*a group $H$ of order $2$, $H=Aut(N)$.

*and a morphism $\phi:G\rightarrow H$.


By 3. and 2. we have that $G/\ker\phi\simeq Im\, \phi$ has order $1$ or $2$ since it's (isomorphic to) a subgroup of $H$.
So your problem is equivalent to checking the statement
$$
G/\ker \phi \simeq H
$$
or equivalently
$$
\phi\text{ is not trivial}.
$$
Group structure of $Aut(N)$
Let $N$ be a group. As a set
$Aut(N):=\{N\rightarrow N, \text{group homomorphism}\}$
The binary operation given by composition
$$
\begin{array}{ccc}
Aut(N)\times Aut(N)&\rightarrow &Aut (N)\\
(F,G)&\mapsto &F\circ G.
\end{array}
$$
Notice that the unit element is the identity map $Id(x) = x$, for every $x\in N$.
Definition of $\phi$
Let $g\in G$, since $N$ is normal subgroup of $G$ then $gxg^{-1}\in N$ for every $x\in N$.
Fix an elment $g\in G$. Check that the map
$$
\begin{array}{ccc}
F_g:&N&\rightarrow &N\\
&x&\mapsto  &gxg^{-1}
\end{array}
$$
is a group homomorphism, i.e. $F_g\in Aut(N)$.
Consider the map 
$$
\begin{array}{ccc}
\phi: &G &\rightarrow &Aut(N)\\
  &g &\mapsto &F_g.
\end{array}
$$
Check that $\phi$ is a group homomorphism. I.e. check that $F_{gh}  = F_g\circ F_h$.
Finishing the proof
Notice that $\phi$ is trivial if and only if $N\subseteq Z(G)$. 
A: Since $N$ is normal, conjugation by $g\in G$ induces an automorphism of $N$, so we obtain a homomorphism $\varphi\colon G\to\operatorname{Aut}(N)$.
Since $N$ is not contained in the center, there is $g\in G$ so that the induced automorphism is not the identity. Since $\operatorname{Aut}(N)$ has two elements, $\varphi$ is surjective and therefore $\ker\varphi$ has index $2$.
The fact that $\ker\varphi=C_G(N)$ is true, but irrelevant.
