Let $G$ be a group, and let $S_G$ be the permutation group of the underlying set $G$ For each $g\in G$ define $\rho_g : G\rightarrow G$ by $\rho_g (h) = gh$ Then $\rho_g$ is invertible with inverse $\rho_{g^{-1}}$ and so is a permutation of the set $G$

Define $\Phi :G\rightarrow S_G$ by $\Phi (g) = \rho_g$ Then $\Phi$ is a homomorphism, since $$(\Phi (gh))(x) = \rho_{gh}(x) = ghx = \rho_g(hx) = (\rho_g\circ\rho_h)(x) = ((\Phi (g))(\Phi (h)))(x)$$

And $\Phi$ is injective, since if $\Phi (g) = \Phi (h)$ then $\rho_g = \rho_h$ so $gx = hx$ for all $x\in X$ and so $g=h$ as required.

So $\Phi$ is an embedding of $G$ into its own permutation group. If $G$ is finite of order $n$ then simply numbering the elements of $G$ gives an embedding from $G$ to $S_n$