Proofs for Multiplicative Functions in Number Theory Part 1:
A theorem in my book proves that if $f(n)$ is a multiplicative function, and $g(n)=\sum_{d|n}f(d)$, then $g(n)$ is also multiplicative. How do I prove the converse of this. The converse being that if $g(n)$ is a multiplicative function, and $g(n)=\sum_{d|n}f(d)$, then $f(n)$ is multiplicative.
Part 2:
This second part goes hand-in-hand with part 1. How do I prove that if $g(n)$ is a multiplicative function, then there is exactly one function $f(n)$ for which $g(n)=\sum_{d|n}f(d)$, and $f(n)$ is a multiplicative function.
What I thought:
For part 1, it looks to me that I might have to use Möbius inversion. I'm not sure though. For part 2, it looks to me that I may have to use induction to prove it. 
 A: Much of this falls under one lump idea. Let 
$$f \star g(n) = \sum_{d \mid n} f(d)g(n/d)$$
be the convolution of $f$ and $g$. Then it happens that if both $g$ and $f \star g$ are multiplicative, then $f$ is also multiplicative. Let $h = f \star g$, for slightly easier notation. Let's prove that now.
We proceed by contradiction. Suppose that $f$ is not multiplicative. So there exist two positive integers $n,m$ with $\gcd(m,n) = 1$ such that $f(mn) \neq f(m)f(n)$. Of all such pairs of integers, let's choose the pair with the smallest product $mn$.
If $m = n = 1$, then $f(1) \neq 1$. Then $h(1) = f(1)g(1) \neq 1$, and so $h$ is not multiplicative. That's a contradiction, and so $f(1) = 1$.
Now for all $a,b$ with $\gcd(a,b) = 1$ and $ab < mn$, we have $f(ab) = f(a)f(b)$. Then writing out the definition of $h(mn)$ but splitting off the factor corresponding to $mn$, we have
$$ h(mn) = \sum_{\substack{a \mid m \\ b \mid n \\ ab < mn}} f(ab)g\left( \frac{mn}{ab}\right) + f(mn)g(1) = \sum_{\substack{a \mid m \\ b \mid n \\ ab < mn}} f(a)f(b) g\left(\frac{m}{a}\right)g\left(\frac{n}{b}\right) + f(mn). $$
We used multiplicativity to get to the right. We can now separate the sums on the right. We'll add and subtract $f(m)f(n)$ too. So the above is equal to
$$ \sum_{a \mid m} f(a)g(m/a)\sum_{b \mid n}f(b)g(n/b) - f(m)f(n) + f(mn) = h(m)h(n) - f(m)f(n) + f(mn).$$
We know that $h$ is multiplicative, so we must have that $h(mn) = h(m)h(n)$, which forces $f(m)f(n) - f(mn) = 0$. Or rather, since I phrased this as a contradiction, we see that $h(mn) \neq h(m)h(n)$, which is a contradiction.
And so $f$ is multiplicative after all. $\diamondsuit$
In your first question, you are looking at $f \star 1$, and both $1$ and $f \star 1$ are multiplicative. So $f$ is multiplicative.
In your second question you can use Möbius inversion. In particular, given $g(n)$, you can consider the function
$$f(n) = \sum_{d \mid n} g(d)\mu(n/d). \tag{1}$$
Then by Möbius inversion, you know that
$$ g(n) = \sum_{d \mid n} f(d).$$
So we have found one such function. You must now show it is unique. Since $g$ is multiplicative, it suffices to consider only prime powers. But the difference in $(1)$ between $g(p^k)$ and $g(p^{k+1})$ is exactly $f(p^{k+1})$. So you can guarantee the values of $f$ are unique at each prime power inductively, starting with $f(1) = 1$.
More generally, you can show that multiplicative functions form a group under $\star$. Alternatively, you can view these as the coefficients of Dirichlet series. Then the corresponding idea is that Dirichlet series that agree at all points have the same coefficients.
A: For part I when you use Mobius inversion you get :
$$f(n)=\sum_{d|n}\mu(n/d)g(d) $$
Now because $\mu$ and $g$ are both multiplicative, $f$ will be multiplicative as well. This is a slightly more general version of your property that :
$$n\mapsto\sum_{d|n}h(n/d)g(d) $$
is a multiplicative function when $h$ and $g$ both are.
For part II there is no need for induction once you use Moebius inversion formula. If $f_1$ and $f_2$ are two functions such that :
$$\sum_{d|n}f_2(d)=g(n)=\sum_{d|n}f_1(d) $$
Then you get that :
$$ f_2(n)=\sum_{d|n}\mu(n/d)g(d)=f_1(n)$$
