LagrangeEuler said:
Well, this is a representation of ##Z_2## group. So it needs to be a group. If I write just Caley table it is confusing. Because of that, I asked the question. Your answers till now were not helpful.
There is no such thing as a Cayley table for a representation.
The Cayley table for ##\mathbb{Z}_2## is
\begin{array}{l|*{5}{l}}
+ & 0 & 1 \\
\hline
0 & 0 & 1 \\
\hline
1 & 1 & 0 \\
\end{array}
As I said earlier, a representation is a group homomorphism ##D\, : \,\mathbb{Z}_2\longrightarrow \operatorname{Aut}(S)## with a set ##S##. Let's assume for simplicity that ##S## is finite, say ##S=\{s_1,\ldots,s_n\}.## Then ##D(0)=\operatorname{id}_S=1## so ##D(0)(s_k)=s_k##. ##D(1)=\sigma## is then a permutation of ##S## and since ##\sigma^2=\sigma\cdot\sigma=D(1)\cdot D(1)=D(1+1)=D(0)=1## it is of order two. The only table I see here is to write
\begin{align*}
D\, : \,\mathbb{Z}_2 &\longrightarrow \operatorname{S}_n\\
0&\longmapsto \operatorname{id}_S = (s_k\longmapsto s_k)\\
1&\longmapsto \sigma=(s_k\longmapsto \sigma(s_k))
\end{align*}
I have the impression that you identified ##0\in \mathbb{Z}_2## with ##\operatorname{id}_S## and ##1\in \mathbb{Z}_2## with ##\sigma\in \operatorname{S}_n## but why? They are in different groups and as such different elements. Since the kernel of ##D## is a subgroup of ##\mathbb{Z}_2##, we don't have many choices. Either this kernel is ##\{0\}## in which case we have a faithful representation and ##\operatorname{Aut}(S)=S_n \trianglerighteq D(\mathbb{Z}_2)\cong \mathbb{Z}_2,## or we have the trivial representation ##D(0)=D(1)=\operatorname{id}_S##.
