Inverting $\displaystyle\sum_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$ into $\displaystyle\sum_{d|n} \lambda(n/d) 2^{\omega(d)}=1$ ,where $n \geq1$, by using Mobius Inversion Formula.
I'm able to solve the latter without Inversion, and in problem too it's not necessary to use inversion, but I'm fascinated to know how to do that, because $2^{\omega(n)}$ comes to L.H.S side from R.H.S and becomes $2^{\omega(d)}$, also $\lambda(d)$ is converted into $\lambda(n/d)$ Please help. and if it's not possible then comment, I'll remove this problem from MSE.