Let $\phi$ be the Euler totient function and let $\tau$ count the divisors of a number. I am interested in the following question: What is the minimum and maximum of $\phi(d)\cdot\tau(n/d)$ where $d$ runs through the divisors of $n$? Is there a "nice" description in terms of $n$? Or maybe if it is easier, I am also interested in the value of the function: $f(n):= \sum_{d|n} \phi(d)^2\cdot \tau(\frac{n}{d})^2$ in terms of the prime factorization of $n$. Here is a small list in case someone recognizes the numbers:
n min max
1 1 1
2 1 2
3 2 2
4 2 3
5 2 4
6 2 4
7 2 6
8 3 4
9 3 6
10 2 8
11 2 10
12 4 6
13 2 12
14 2 12
15 4 8
16 4 8
17 2 16
18 3 12
19 2 18
20 4 12
Let $p \mid n$ be prime, $a$ its exponent in $n$ and $0\leq b \leq a$ be its exponent in $d \mid n$.
We want to know when $$f(b) :=\phi(p^b)\tau(p^{a-b}) = \begin{cases}(p-1)p^{b-1}(a-b+1) &: b >0 \\ a+1 &: b=0\end{cases}$$ is maximal and minimal.
For maximal: We have $m+1 \leq 2^m$ (*) for $m=a-b \in \mathbb N$ so $f(b) \leq f(a)$ for $b>0$. For $b=0$ we have $f(0) \leq f(a)$ unless $p=2$ and $a \in \{1,2\}$. In those cases, $f(0,1,2) = a+1, a, 2$, and the maximum is attained for $b=0$.
Conclusion: $\phi(d)\tau(n/d)$ is maximal for $d=n$ if $n$ is odd or divisible by $8$; otherwise, for $d$ equal to the odd part of $n$.
Looking more carefully at the inequalities above, we see that all values of $d$ where the maximum is attained are:
For minimal: For $0<b$ be we have $a+1 \leq 2^b(a-b+1)$ and even $a+1 \leq 2^{b-1}(a-b+1)$ unless $b=1$ (**) Hence for $p>2$, $f(b)$ is minimal for $b=0$. For $p=2$, the minimum is for $b=1$.
Conclusion: $\phi(d)\tau(n/d)$ is minimal for $d=1$ if $n$ is odd, and $d=2$ if $n$ is even.
(*) If $S$ is a set with $m$ elements, it embeds in its power set by sending each element to the corresponding singleton. There is at least one additional element in the power set: the empty set.
(**) For $0<b\leq a$, a sequence with $a$ elements embeds into a set with $2^b (a-b+1)$ elements: the latter counts for each (consecutive) subsegment of length $b$ the number of subsets of that subsegment. We can again send each element of the sequence to a singleton. There is much place left ($b-a+1$ empty sets and then some). Similarly with the exponent $2^{b-1}$, by using the inequality $b \leq 2^{b-1}$ for the last segment of length $b$ and $1 \leq 2^{b-1}$ for all others. In that case, there is again one element left unless $b=1$.