So the question is "What are all possible actions by automorphism of H on N?" with H = Z/3Z and N = Z/6Z.
I completely guessed my way through how to go about solving this, but I started with finding Aut(Z/6Z) which is {[1],[5]}, and figured that it Aut(Z/6Z) must be isomorphic to Z/2Z. Working through I found that the only possible action is the direct action Z6 x Z/3Z.
This is most likely wrong, and any help will be greatly appreciated.
As you've computed, the group of automorphisms of $\mathbb{Z}/6\mathbb{Z}$ is of order $2$. An action of $\mathbb{Z}/3\mathbb{Z}$ on $\mathbb{Z}/6\mathbb{Z}$ thus arises from a homomorphism from a group of order 3 into a group of order 2, and any such homomorphism is trivial. Thus there is only the trivial action.