Every group $G$ is a subgroup of $\operatorname{Sym}(G)$ I am not sure why this fact makes sense.
Say $G = \{e, a, a^2\}$
The order of the elements of $G$ are irrelevant in $G$ itself. So what does $G$ look like in the 6 elements of $\operatorname{Sym}(G)$?
 A: Note that $G$ is not a subgroup of (in this case) the symmetric group on $\{e,a,a^2\}$. But $G$ is in a nice way isomorphic to a subgroup of the symmetric group.
The element $e$ is of course taken to the identity permutation. 
The element $a$ is taken to the permutation that takes $e$ to $a$, $a$ to $a^2$, and $a^2$ to $e$.
Finally, $a^2$ is taken to the permutation that takes $e$ to $a^2$, $a$ to $e$, and $a^2$ to $a$. 
Remark: In general, the element $g$ is taken to the permutation $\pi_g$ of $G$ obtained by multiplying  elements of $G$, say on the right, by $g$. It is easy to verify that $\pi_g \pi_h=\pi_{gh}$ for all $g$, $h$ in $G$. This gets us most of the way to proving the isomorphism result.
A: You can embed every group into $S_n$ where $n=|G|$.  For example, your group $G=C_3$ can embed into $S_3$ by the mapping $a\mapsto (123)$.
But since you're looking for an explanation specifically into $\mbox{Sym}(G)$ (which by the way is isomorphic to $S_n$), we can define $\sigma_g:G\rightarrow G$ by $\sigma_g(h)=hg$, and then let $\sigma:G\rightarrow \mbox{Sym}(G)$ be defined by $\sigma(g)=\sigma_g$.  (This is called the "right regular representation" of $G$.)  It's pretty easy to verify this is a homomorphism.  Now, if $g\in \mbox{Ker}(\sigma)$, $\sigma_g$ is the trivial permutation (that is, $hg=\sigma_g(h)=h$ for all $h\in G$).  But this implies that $g=e$, so $\sigma$ is injective.  By the first isomorphism theorem $G$ is isomorphic to its image which is a subgroup of $\mbox{Sym}(G)$.
So by this construction, you can see that $G=C_3$ as you've defined it is isomorphic to $\{\sigma_e,\sigma_a,\sigma_{a^2}\}$ as defined above.  Does this help?
