This comes from Aluffi's Algebra: Chapter $0$. The the first part of the question was to prove for $n \geq 5$, show that there are no nontrivial actions of $A_{n}$ on any set $S$ with $|S| < n$. The next part was to construct a nontrivial action of $A_{4}$ on a set $S$ with $|S| = 3$.
An action of $A_{4}$ on a $3$-set is the same as a homomorphism $\varphi:A_{4} \to S_{3}$. As $|A_{4}| = 12$ and $|S_{3}| = 6$, $\varphi$ cannot be injective. $A_{4}$ contains a normal subgroup of order $4$, consisting of all permutations of type $[2,2]$ with identity, and this must be the kernel of $\varphi$ as $\varphi$ is not injective and nontrivial. By the first isomorphism theorem, the image of $\varphi$ must consist of three elements of $S_{3}$. My first thought was something like this $$\varphi(\sigma) = \begin{cases} \text{Id}_{S_{3}} & \sigma = \text{Id}_{A_{4}}, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3) \\ (1 \ 2 \ 3) & \sigma = (1 \ 2 \ 3), (1 \ 2 \ 4), (1 \ 3 \ 4), (2 \ 3 \ 4) \\ (1 \ 3 \ 2) & \sigma = (1 \ 3 \ 2), (1 \ 4 \ 2), (1 \ 4 \ 3), (2 \ 4 \ 3) \end{cases}$$ But this is not a homomorphism. Is there a better way at constructing such a homomorphism than guessing and checking?
The action $\varphi$ by left multiplication of $A_4$ on the set of the three cosets of $V_4=\{(),(12)(34),(13)(24),(14)(23)\}\le A_4$ is not trivial, because its kernel is: $$\ker\varphi=\bigcap_{\sigma\in A_4}\sigma V_4\sigma^{-1}\le V_4<A_4$$