On the representable functors page on Wikipedia, they mention that, if we consider a group $G$ as a category with a single object $\star$ and all arrows invertible, we can view a $G$-action on a set as a functor $F: G \to \mathsf{Set}$. Suppose $F(\star) = S$. Then $F$ is representable iff there is a natural bijection between $S$ and $G(\star, \star)$, i.e. the elements of the group.
The article also mentions that an action $F$ admits a representation iff the action is simply transitive, in which case a representation amounts to a selection of an element $s \in S$, and a correspondence (bijection) $\operatorname{Ar}(G) \to S: g \mapsto g \cdot s$.
My question is: this seems to bring about a natural bijection
$$\text{Natural isomorphisms } F \Rightarrow G(\star, -) \quad \longleftrightarrow \quad \text{Elements of $S$},$$
which reminds one of the Yoneda Lemma. But shouldn't the Yoneda Lemma give us a bijection between all natural transformations and elements of $S$, not just the isomorphisms?
Edit: The natural transformations should go rather from $G(\star, -) \Rightarrow F$.
I think I answered my own question: the existence of any natural transformation requires $F$ to be a simply transitive group action, in which case I claim any natural transformation $G(\star, -) \Rightarrow F$ must be an isomorphism.