Divisor sum property of Euler phi function with Mobius inversion
In the property for Euler's totient function we just use that the divisors $d$ are in bijection with the codivisors $n/d$, so that $$\sum_{d|n}\phi\left(\frac{n}{d}\right) = \sum_{d|n}\phi(d)=n$$ This is also called Sum Over Divisors Equals Sum Over Quotients.
However, the Moebius inversion formula, for $f(n)=\phi(n)$ gives $g(n)=\sum_{d\mid n}f(d)=n$, and not $g(n)=\phi(n)$. So this is impossible.