Two actions of $\mathbb{Z}_6$ on itself that we naturally might consider are $\overline{m} \cdot \overline{n} = \overline{m}+\overline{n}$, and $\overline{m} \cdot \overline{n} = \overline{n}-\overline{m}$. In fact these actions are isomorphic. On the other hand, we can define an action such as $\overline{m} \cdot \overline{n} = 2\overline{m} + \overline{n}$, which is not isomorphic to those.
So my question is whether all the transitive actions of $\mathbb{Z}_6$ acting on itself are isomorphic to the above $\overline{m} \cdot \overline{n} = \overline{m}+\overline{n}$.
Suppose that $\mathbb{Z}_6$ acts on itself in some way. Then, in order for the action to be transitive, the map $\bar{m} \mapsto \bar{m} \cdot \bar{0}$ (which is easily seen to be an equivariant map, if the codomain has the given action and the domain has the "natural" action given by addition) must be surjective. Since it is a surjective map from a finite set to itself, it must also be injective, and hence an isomorphism of $\mathbb{Z}_6$-sets.