TITLE: Do these groups have a meaning?
QUESTION [7 upvotes]: Let $G$ be a group. We can say that $Aut(G)\leq S_G$ where $S_G$ denotes the set of all bijection from $G$ to $G$.
But $S_G$ is not a good bound for $Aut(G)$ as $S_G$ grows very fast. Let's define new groups.
$$C(G)=\{f\in S_G|f(x^k)=f(x)^k \text{for all $x\in G $ and $k \in \mathbb Z$}  \}$$ 
$$A(G)=\{f\in S_G|f(xy)=f(x)f(y) \text{if $xy=yx$}  \}$$
It is easy to see that $C(G)$ and $A(G)$ are both subgroups of $S_G$ and we can see that
$$Aut(G)\leq A(G)\leq C(G)\leq S_G$$
Are these groups good bound for $Aut(G)$  or are they important for determining the structure of $G$ ?
Edit: One simple observation; If $G$ is abelian then $Aut(G)=A(G)$ and if $G$ is cyclic then $Aut(G)=C(G)$.
Edit$2$: Ahulpke provides a nice table for $A(G),C(G)$ and $Aut(G)$. In the table, for some groups, $|A(G):Aut(G)|=2$ so $Aut(G)$ is normal in $A(G)$. Is $Aut(G)$ always normal in $A(G)$ ?

REPLY [3 votes]: As $1^2=1$, the group $C(G)$ must stabilize the identity, and then in turn stabilize the partition of $G$ by element orders.
The following table gives, for $G$ taken from small groups, the orders of order stabilizer, $C(G)$, $A(G)$ and $Aut(G)$. Clearly all groups can be different, sometimes notably so.
If the group id is underlined, then $Aut(G)$ is normal in $A(G)$ (otherwise not).
\begin{equation}
\begin{array}{l|l|r|r|r|r}
Id
&Name
&Ord
&C(G)
&A(G)
&Aut(G)
\\
\hline
\underline{s4n2}
&C2 x C2
&2
{\cdot}3
&2
{\cdot}3
&2
{\cdot}3
&2
{\cdot}3
\\
\underline{s6n1}
&S3
&2^{2}3
&2^{2}3
&2^{2}3
&2
{\cdot}3
\\
\underline{s8n2}
&C4 x C2
&2^{4}3^{2}
&2^{4}
&2^{3}
&2^{3}
\\
\underline{s8n3}
&D8
&2^{4}3
{\cdot}5
&2^{4}3
&2^{4}
&2^{3}
\\
\underline{s8n4}
&Q8
&2^{4}3^{2}5
&2^{4}3
&2^{4}3
&2^{3}3
\\
\underline{s8n5}
&C2 x C2 x C2
&2^{4}3^{2}5
{\cdot}7
&2^{4}3^{2}5
{\cdot}7
&2^{3}3
{\cdot}7
&2^{3}3
{\cdot}7
\\
\underline{s9n2}
&C3 x C3
&2^{7}3^{2}5
{\cdot}7
&2^{7}3
&2^{4}3
&2^{4}3
\\
s10n1
&D10
&2^{6}3^{2}5
&2^{5}3
{\cdot}5
&2^{5}3
{\cdot}5
&2^{2}5
\\
s12n1
&C3 : C4
&2^{6}3^{2}5
&2^{5}3
&2^{5}3
&2^{2}3
\\
s12n3
&A4
&2^{8}3^{3}5
{\cdot}7
&2^{8}3^{2}
&2^{8}3^{2}
&2^{3}3
\\
s12n4
&D12
&2^{6}3^{2}5
{\cdot}7
&2^{5}3^{2}5
&2^{5}3
&2^{2}3
\\
\underline{s12n5}
&C6 x C2
&2^{6}3^{3}5
&2^{2}3
&2^{2}3
&2^{2}3
\\
s14n1
&D14
&2^{8}3^{4}5^{2}7
&2^{5}3^{3}5
{\cdot}7
&2^{5}3^{3}5
{\cdot}7
&2
{\cdot}3
{\cdot}7
\\
\underline{s16n2}
&C4 x C4
&2^{11}3^{6}5^{2}7
{\cdot}11
&2^{10}3
&2^{5}3
&2^{5}3
\\
s16n3
&(C4 x C2) : C2
&2^{11}3^{4}5^{2}7^{2}
&2^{10}3
{\cdot}5
&2^{7}
&2^{5}
\\
\underline{s16n4}
&C4 : C4
&2^{11}3^{6}5^{2}7
{\cdot}11
&2^{10}3
&2^{7}
&2^{5}
\\
\underline{s16n5}
&C8 x C2
&2^{11}3^{4}5
{\cdot}7
&2^{6}
&2^{4}
&2^{4}
\\
\underline{s16n6}
&C8 : C2
&2^{11}3^{4}5
{\cdot}7
&2^{6}
&2^{5}
&2^{4}
\\
s16n7
&D16
&2^{11}3^{5}5
{\cdot}7
&2^{9}3^{2}5
{\cdot}7
&2^{9}3
&2^{5}
\\
s16n8
&QD16
&2^{10}3^{4}5^{2}
&2^{8}3
&2^{8}
&2^{4}
\\
s16n9
&Q16
&2^{11}3^{5}5^{2}7
&2^{9}3
&2^{9}3
&2^{5}
\\
\underline{s16n10}
&C4 x C2 x C2
&2^{11}3^{4}5^{2}7^{2}
&2^{11}3^{3}5
&2^{6}3
&2^{6}3
\\
s16n11
&C2 x D8
&2^{11}3^{5}5^{2}7
{\cdot}11
&2^{11}3^{4}5^{2}7
&2^{8}
&2^{6}
\\
s16n12
&C2 x Q8
&2^{11}3^{6}5^{2}7
{\cdot}11
&2^{11}3^{2}5
&2^{8}3
&2^{6}3
\\
\underline{s16n13}
&(C4 x C2) : C2
&2^{11}3^{4}5^{2}7^{2}
&2^{11}3^{3}5
&2^{5}3
&2^{4}3
\\
\underline{s16n14}
&C2 x C2 x C2 x C2
&2^{11}3^{6}5^{3}7^{2}11
{\cdot}13
&2^{11}3^{6}5^{3}7^{2}11
{\cdot}13
&2^{6}3^{2}5
{\cdot}7
&2^{6}3^{2}5
{\cdot}7
\\
\end{array}
\end{equation}