$\displaystyle
n\sum_{d|n} \frac{1}{d} = \sum_{d|n} \frac{n}{d} = \sum_{d|n} {d} = \sigma (n)
$
or
$\displaystyle
\frac{\sigma (n)}{n} = \frac{1}{n} \sum_{d|n} {d} = \sum_{d|n} \frac{d}{n} = \sum_{d|n} \frac{1}{d}
$
$$\sum_{d\mid n} \frac{n}d \sigma(d) =\sum_{d_1\mid n}\frac{n}{d_1}\sum_{d_2\mid d_1} d_2 = \sum_{d_2\mid d_1\mid n}\frac{n}{d_1/d_2}$$
$$\sum_{d\mid n} d\tau(d)= \sum_{d_3\mid n} d_3\sum_{d_4\mid d_3}1 = \sum_{d_4\mid d_3\mid n} d_3$$
Now, map $(d_1,d_2)$ to $(d_3,d_4)=(nd_2/d_1,n/d_1)$ and we see we have the same sums.
So, more generally, if $S_n=\{(d_1,d_2): d_2\mid d_1\mid n\}$ then the map $S_n\to S_n$ defined by $(d_1,d_2)\to\left(\frac{nd_2}{d_1},\frac n{d_1}\right)$ is a bijection. Thus, for any function $f(m,n)$ of two natural numbers, we have that:
$$\sum_{(d_1,d_2)\in S_n} f(d_1,d_2)=\sum_{(d_1,d_2)\in S_n} f\left(\frac{nd_2}{d_1},\frac{n}{d_1}\right)$$
The above is just the case of $f(m,n)=m$.
Best Answer
By definition $\nu(n)=\sum_{d|n} 1$ and $\sigma(n)=\sum_{d|n} d$ for every integer $n>0$ so that $\{\nu,1\}$ and $\{\sigma,\operatorname{id}\}$ (with $1:n\mapsto 1$ and $\operatorname{id}:n\mapsto n$) will be both Möbius pairs according to the Möbius inversion formula.