Let $G$ be a finite group of order $n$. By Cayley's theorem, for any bijection $f\colon G\to X:=\{1,\dots,n\}$, $G$ embeds into $S_n$ as a transitive permutation group. Does the converse hold true? More precisely:
Question. Let $\Sigma=\{\sigma_1,\dots,\sigma_n\}\le S_n$ be a transitive permutation group. Does there always exist a group $G=\{g_1,\dots,g_n\}$ and a bijection $f$ (as above) such that, for every $i,j\in X$: $$\sigma_i(j) = f(g_i(f^{-1}(j))) \space?\tag1$$
Yes, this is true. The general setting is: let $Y$ be a set and $S_Y$ the group of bijections $Y \to Y$, the symmetric group on $Y$, and $G\leq S_Y$ be any subgroup with the same cardinality as $Y$, i.e. $\#G = \# Y$. Let $\chi: G \to S_G, \ g \mapsto l_g$ be the Cayley embedding and $\iota: G \to S_Y, \ g \mapsto g$ the natural inclusion.
Then a bijection $\Phi: G \to Y$ with $$c_\Phi \circ \chi = \iota$$ is needed (where $c_\Phi$ is conjugation). This is equivalent to $$g = c_\Phi\circ \chi(g) = \Phi \circ l_g \circ \Phi^{-1} \ \forall g \in G \iff g(\Phi(h)) = \Phi(gh) \ \forall g,h \in G$$ i.e. $\Phi$ is a $G$-equivariant bijection. In particular, $\Phi$ is determined in this case by a single element $\Phi(1)=y \in Y$, because $\Phi(g) = g\Phi(1)$ for all $g \in G$, i.e. $\Phi = \Phi_y$ with $\Phi_y(g) = gy$ for $g \in G$.
$\Phi_y$ is surjective $\iff Gy =Y \iff G$ is transitive. $\Phi_y$ is injective $\iff G_y = 1,$ i.e. the stabilizer of $y$ is trivial.
Since in the OP case $G$ is finite (and $X =Y$ and $\# X = \# G$) it holds $\Phi_y$ is surjective $\iff$ injective $\iff$ bijective. So, since $G$ is assumed transitive $\Phi$ exists.