Before proving that $Aut(G)$ is a subgroup of $S_G$ ... Exercise 37 Ch. 9 of the book Abstract Algebra by T Judson :

We will denote the set of all automorphisms of G by $Aut(G)$. Prove that 
  $Aut(G)$ is a subgroup of $S_G$ , the group of permutations of G. 

Let $G= \mathbb {Z_6}$. Then there are two isomorphisms between $\mathbb {Z_6}$ and $\mathbb {Z_6}$ with additive operation. One is $\phi (x)=x$ the identity, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{0,1,2,3,4,5}\}$, and the other is $\phi (x)=x+5$, i.e. ${\{0,1,2,3,4,5}\} \longleftrightarrow {\{5,0,1,2,3,4}\}$. But the set ${\{(1), (054321)}\}$ is not a group at all let alone to be a subgroup of $S_{|\mathbb{Z_6}|}=S_6$. 
Where am I wrong?    
 A: The automorphisms of $(\mathbb {Z_6},+)$ are $x \mapsto ax$, for $a=\pm1$.
They form a cyclic group of order $2$, which is a subgroup of $S_6$.
A: Your second map is not an automorphism, because it doesn’t even take the identity ($0$) to itself. Rather, the nontrivial automorphism is $x\mapsto-x$, so leaves $0$ and $3$ fixed, otherwise $1\leftrightarrow5$, $2\leftrightarrow4$.
A: The two automorphisms of $\mathbb {Z_6}$ are:
$\phi_1: \phi_1 (x)=x$ 
$\phi_2: \phi_2 (x)=5x~ mod ~6$
$\phi_1$ is the map (012345) $\longleftrightarrow$  (012345), the identity.
$\phi_2$ is the map (012345) $\longleftrightarrow$  (501234).
These two automorphisms form a group under the composition $\circ$:
$\phi_1 \circ \phi_1 = \phi_1$ 
$\phi_1 \circ \phi_2 = \phi_2 \circ \phi_1 = \phi_2 $ 
$\phi_2 \circ \phi_2 = \phi_1$
which gives the same multiplication table of $\mathbb {Z_2}$ and so $Aut(\mathbb {Z_6}) = \mathbb {Z_2}$
Your $\phi_2 = x+5$ is not correct because if you apply it twice it won't give you the identity. 
An equivalent way to express the automorphisms is:
$\phi_1: \phi_1 (x)=x$ 
$\phi_2: \phi_2 (x)=-x$
which is again $\mathbb {Z_2}$.
