Abelian characters and odd perfect numbers?

Question

This question is about applications of abelian characters to odd perfect numbers:

Context and Definitions:

Let $n$ be a natural number and $D_n$ be the set of divisors. We can make this set to a ring by observing that each divisor $d$ has

$$0 \le v_p(d) \le v_p(n)$$

Hence we can add two divisors $d,e$ by setting:

$$d \oplus e := \prod_{p | n} p^{v_p(d)+v_p(e) \mod (v_p(n)+1)}$$

and similarily we can multiply them by setting: $$d \otimes e := \prod_{p | n} p^{v_p(d) \cdot v_p(e) \mod (v_p(n)+1)}$$

Then, if $n = p_1^{a_1} \cdots p_r^{a_r}$, this ring will be isomorphic to the ring

$$\mathbb{Z}/(a_1+1) \times \cdots \times \mathbb{Z}/(a_r+1)$$

If $n$ is squarefree, than this reduces to :

$$d\oplus e = \frac{de}{\gcd(d,e)^2}$$

$$d\otimes e = \gcd(d,e)$$

Let $\phi_n(a):= \sum_{p|a} \frac{v_p(a)}{\sqrt{v_p(n)+1}}e_p$, then:

$$ \langle \phi_n(a),\phi_n(b) \rangle = \sum_{p|\gcd(a,b)} \frac{v_p(a)v_p(b)}{v_p(n)+1} =:K_n(a,b)$$

Set

$$\chi_n(d,e) := \chi^{(n)}_d(e):=\chi^{(n)}_e(d):=\exp\left(2 \pi \sqrt{-1} K_n(d,e) \right )$$

. Fix some natural number $n$ and let $\chi_d=:\chi^{(n)}_d$:

$$C(n) := \{ d:d|n, \chi_d = \bar{\chi}_d \}$$

These (two) questions came after researching the set $C(n)$ empirically with SageMath:

1. $C(d) \cdot C(\frac{n}{d}) =^? C(n) \forall d \in \mathbf{U}(n):= \{d:d|n, \gcd(d,n/d)=1\}$

2. $C(p^a) =^? \{1\}$ if $ a \equiv 0 \mod(2)$ and $= \{1, p^{\frac{a+1}{2}} \}, , a \equiv 1 \mod(2)$

3. Follows from 1) and 2):

$$C(n) = \prod_{p|n, v_p(n) \equiv 1 \mod(2)}C(p^{v_p(n)})=\prod_{p|n, v_p(n) \equiv 1 \mod(2)} \{ 1, p^{\frac{v_p(n)+1}{2}} \}$$

4. Follows from 3):

$$C(n)=\{ \sqrt{d \operatorname{rad}(d)}: d|n, \gcd(d,n/d)=1, \forall p| d : v_p(d) \equiv 1 \mod(2) \}$$

5. Follows from 4): For an odd perfect number in Euler form $n = q^a m^2, q \equiv a \equiv 1 \mod(4),q$ prime, it follows that $C(n)=\{1,q^{\frac{a+1}{2}}\}$

Odd perfect numbers:

Let now $n$ denote an odd perfect number and $e:=q^{\frac{a+1}{2}}$ and consider $\chi_e = \bar{\chi}_e$. With this character, we can divide the set $D_n$ into two subsets:

$$D_+(n) \cup D_-(n) = D_n$$

defined through: $D_\pm(n):= \{d \in D_n| \chi_e(d) = \pm 1\}$. The interesting part is now that there are two bijections on these sets given by:

$$\alpha: D_+(n) \rightarrow D_-(n) : d^+ \mapsto \frac{n}{d^+}$$

and

$$\beta: D_+(n) \rightarrow D_-(n) : d^+ \mapsto q d^+$$

Meta-Question: Can there be said anything new relating odd perfect number using this character and these two bijections? (I have done some calculations along these lines and wanted to see if there are some other ideas relating these issues.)

Are there other usages of abelian characters to odd perfect numbers?

Thanks for your help!