A possible axiomatic characterization of the set of divisors of a perfect number

Question

Define a pro-perfect set $S$ to be a finite set of positive integers satisfying the following three properties:

1. $1\in S$.
2. $\displaystyle\sum_{n\in S}n^{-1}\in S$
3. There exists a unique permutation $\pi$ of $S$ of order $2$ such that $$\frac{1}{\vert S\vert}\sum_{n\in S}n\pi(n)\in S.$$

Question 1: Is any pro-perfect set necessarily the set of divisors of a perfect number?

Question 2: Does every pro-perfect set necessarily contain $2$?

Edit 1: it might help to consider the more general notion of $k$-pro-perfect set such that property 3 is replaced by

3' : There exists a (not necessarily unique) permutation $\pi$ of order $k$ such that $$\frac{1}{\vert S\vert}\sum_{n\in S}\prod_{\ell=0}^{k-1}\pi^{\circ \ell}(n)\in S.$$

Edit 2: Let $s_{\text{inv}}$ denote the quantity $$s_{\text{inv}}:=\frac{1}{\vert S\vert}\sum_{n\in S}n\pi(n),$$ one gets the following equivalence: $$\forall n\in S, \quad n\mid s_{inv}\;\Longleftrightarrow\;\pi:n\mapsto\frac{s_{\text{inv}}}{n}.$$

Edit 3: let $s_{rec}$ denote $\sum_{s\in S}n^{-1}$ and suppose the $m$ first elements of $S$ when arranged in increasing order form a geometric progression with constant ratio $q$. Then $s_{rec}\geq\frac{1-q^{-m}}{1-q^{-1}}$. The latter is less than $2$ as $q\geq 2$. Moreover $s_{rec}\leq\frac{\vert S\vert(1-q^{-m)}}{m(1-q^{-1})}<\frac{2\vert S\vert}{m}$.

Answer 1

Try this.

$$ S = \{1,2,3,12,18,36\} $$ then $$ \sum_{k \in S}\frac{1}{k} = 2 \in S $$ and $$ \frac{1}{6}\sum_{k \in S} k \pi(k) = 36 \in S $$ when $\pi$ is the permutation of $S$ that reverses the order, while $$ \frac{1}{6}\sum_{k \in S} k \pi(k) > 36 $$ for any other permutation $\pi$.

Uniqueness of the minimizing permutation is from Hardy-Littlewood-Polya rearrangement.

Answer 2

Note: $\sigma(n) = 3n$ for $n=120$.
[$\sigma$ the sum of the divisors]

Let $S = \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$ be the set of all positive divisors of $120$. Then $|S| = 16$, $$ \sum_{k \in S}\frac{1}{k} = 3 \in S $$ and $$ \frac{1}{16}\sum_{k \in S} k\pi(k) = 120 \in S $$ where $\pi$ is the permutation that reverses the order, and $$ \frac{1}{16}\sum_{k \in S} k\pi(k) > 120 $$ for all other permutations.

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Note: if $S$ is the set of all divisors of some positive integer $m$, then $$ \sum_{k \in S} \frac{1}{k} = \frac{\sigma(m)}{m} $$ and $$ \frac{1}{|S|}\sum_{k \in S} k\pi(k) = m $$ where $\pi$ is the permutation that reverses $S$, and $$ \frac{1}{|S|}\sum_{k \in S} k\pi(k) > m $$ for all other permutations.