The multiplication of the default ring structure gives a group action of $(\mathbb Z/n \mathbb Z)^*$ on $\mathbb Z/n \mathbb Z$: $$(\mathbb Z/n \mathbb Z)^* \oplus \mathbb Z/n \mathbb Z \ni (a, m) \mapsto a \cdot m \in \mathbb Z/n \mathbb Z$$
Therefore, we have a first cohomology group $H^1((\mathbb Z/n \mathbb Z)^*, \mathbb Z/n \mathbb Z)$.
After some calculating, which took about a week of spare time, I got to the following result: $$ \#H^1((\mathbb Z/n \mathbb Z)^*, \mathbb Z/n \mathbb Z) = \begin{cases} 1 & \text{if}\ 2 \nmid n \\ 2^{\{\text{prime factors of n}\} - 1} & \text{if}\ 2 \mid n, 4 \nmid n \\ 2^{\{\text{prime factors of n}\}} & \text{if}\ 4 \mid n \end{cases}. $$
Is this correct?
I do not think that I am the first to have done this, but in my initial search, I was not really able to find any literature on it. Is there any literature on $H^1((\mathbb Z/n \mathbb Z)^*, \mathbb Z/n \mathbb Z)$? Somewhere where I could have looked up this result?