Here is a sample result for the function, essentially classifying its uniqueness :

Proposition 1: $\mu$ is the unique mapping $\Nstar\to\Z$ such that \begin{eqnarray} \mu(1)&=&1 \\ \sum_{d|n}\mu(d)&=&0\textrm{ for all }n>1 \end{eqnarray} Proof: By induction, there can only be one function with these properties. $\mu$ clearly satisfies (1), so take some $n>1$ . Let $p$ be some prime factor of $n$ , and let $m$ be the product of all the prime factors of $n$ . \begin{eqnarray*} \sum_{d|n}\mu(d)&=&\sum_{d|m}\mu(d) \\ &=&\sum_{\substack{d|m\\p\nmid d}}\mu(d) +\sum_{\substack{d|m\\p\mid d}}\mu(d) \\ &=&\sum_{d|m/p}\mu(d)-\sum_{d|m/p}\mu(d) \\ &=&0 \end{eqnarray*}