Find all $f$ such that $\sum_{d|n}f(d)=\varphi(n)$ The following is the problem:

Let $f:\mathbb{N}\rightarrow \mathbb{C}$ be given by $$\sum_{d|n} f(d)=\varphi(n)$$
  for all $n$, where $\varphi$ is the Euler Totient Function. Find all such $f$. 

To begin with, I do not really understand what the problem is asking, what does it mean to find all such $f$? Isn't $f$ already given? (or that I am unsure how the answer is supposed to look like, in what type of description is it supposed to be). In any case, I have a feeling that the Mobius Inversion will be involved, so by Mobius Inversion, we can say $$f(n)=\sum_{d|n}\mu(d)\varphi(\frac{n}{d})$$ for all $n$. where $\mu$ is the Mobius Function. Is this enough? (I have a feeling it is not), if not, how can I proceed? I appreciate any help. Thank you.
 A: The problem is asking for the arithmetic functions $f:\mathbb{N}\to\mathbb{C}$ such that $(f*1)=\varphi$, where $*$ is Dirichlet's convolution. By Moebius inversion formula the solution is unique, and it is given by $f=\varphi * \mu$, i.e.
$$ f(n) = \sum_{d\mid n} \mu(d)\,\varphi\left(\tfrac{n}{d}\right) $$
which is a multiplicative function such that
$$ f(p) = \varphi(p)-\varphi(1) = p-2 = p\left(1-\frac{2}{p}\right), $$
$$ f(p^k) = \varphi(p^k)-\varphi(p^{k-1}) = p^k \left(1-\frac{1}{p}\right)^2 $$
for any prime $p$ and any $k\geq 2$. It follows that
$$ f(n) = n\prod_{p|| n}\left(1-\frac{2}{p}\right)\prod_{p:p^2\mid n}\left(1-\frac{1}{p}\right)^2. $$ 

An alternative approach is to notice that the Dirichlet series associated with $\varphi$ is $\frac{\zeta(s-1)}{\zeta(s)}$, hence the Dirichlet series associated with $f$ has to be 
$$\frac{\zeta(s-1)}{\zeta(s)^2}=\sum_{n\geq 1}\frac{f(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)^2 \left(1-\frac{1}{p^{s-1}}\right)^{-1}.$$
The conclusion is clearly the same.
