Given $$f(n)=\frac1n\sum_{\substack {a=1\\(a,n)=1}}^{n}a$$ Now I want to show that it satisfies $$\sum_{d|n}f(d)=\frac12(n+1)$$ And then use it to show that for $n>1$: $$f(n)=\frac12\phi(n)$$ Note: Can't use $f(n)=\phi(n)/2$ derived using pairs $(a,n-a)$.
Now this is what I did: $$\begin{align} \sum_{d|n}f(d)&=\sum_{d|n}\frac1d\sum_{\substack {a=1\\(a,d)=1}}^{d}a\\ &=\sum_{d|n}\frac1d\sum_{a=1}^da\sum_{d'|(a,d)}\mu(d')\\ &=\sum_{d|n}\frac1d\sum_{a=1}^da\sum_{\substack{d'|a\\d'|d}}\mu(d')\\ &=\sum_{d|n}\sum_{d'|d}\sum_{m\le d/d'}\frac {d'm}{d}\mu(d')\\ &=\sum_{d|n}\sum_{d'|d}\mu(d')\frac {d'}d\sum_{m\le d/d'}m\\ &=\sum_{d|n}\sum_{d'|d}\mu(d')\frac {d'}d\left(\frac12\frac d{d'}\left(1+\frac d{d'}\right)\right)\\ &=\frac12\sum_{d|n}\left[\sum_{d'|d}\mu(d')+d\sum_{d'|d}\frac{\mu(d')}{d'}\right]\\ &=\frac12\sum_{d|n}(\nu(d)+\phi(d))\\ &=\frac12\nu(1)+\frac12\sum_{\substack{d|n\\d\ne 1}}\nu(d)+\frac12\sum_{d|n}\phi(d)\\ &=\frac12\times 1+\frac12\times0+\frac12\times n\\ &=\frac12(n+1) \end{align}$$ Further using Möbius inversion formula: $$g(n)=\frac12(n+1)=\sum_{d|n}f(d)\iff f(n)=\sum_{d|n}\frac12\left(\frac nd+1\right)\mu(d)$$ SO: $$\begin{align}f(n)&=\frac n2\sum_{d|n}\frac{\mu(d)}d+\frac 12\sum_{d|n}\mu(d)\\ &=\frac n2\phi(n)+\frac12\nu(n)=\frac n2\phi(n)\tag{$\because n>1$}\end{align}$$