Does $\sigma(n^2)/q \mid q^k n^2$ imply $\sigma(n^2)/q \mid n^2$, if $q^k n^2$ is an odd perfect number with special prime $q$? Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.  Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
I discovered an interesting identity involving divisors of odd perfect numbers given in the Eulerian form $N = q^k n^2$ today (July 13, 2021).  (Recall that an odd perfect number $N = q^k n^2$ is said to be given in Eulerian form if $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)
The identity is:

Proposition:  If $N = q^k n^2$ is an odd perfect number with special prime $q$, then
$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$

Proof:
Our starting point is the following blog post, where it is proved that
$$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$
However, note that we have
$$\frac{\sigma(n^2)}{q^k} = \gcd(n^2, \sigma(n^2)).$$
These equations are equivalent to
$$2(1 - q)n^2 + q\sigma(n^2) = \frac{\sigma(n^2)}{q^k}.$$
Factoring out $qn^2$ on the LHS, we obtain
$$qn^2 \Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q^k}.$$
Multiplying both sides of the last equation by $q^{k-1}$, we get
$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$
(Note that the RHS of the last equation is an odd integer.)
This concludes our proof.
QED.
In particular, we have
$$\frac{N}{\sigma(n^2)/q} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)}.$$
But we also know from the following MSE post that
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{2(q - 1)}{q\bigg(q^{k+1} - 1\bigg)}.$$
This means that we obtain
$$\frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\bigg(q^{k+1} - 1\bigg)}{2(q - 1)} = \frac{q\sigma(q^k)}{2}.$$
But $\sigma(q^k) \equiv k + 1 \pmod 4$, since $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$
This finding implies that $\sigma(n^2)/q$ divides $N = q^k n^2$.
Here is my:

QUESTION: Does $\sigma(n^2)/q \mid q^k n^2$ imply that $\sigma(n^2)/q \mid n^2$?  If not, under what condition(s) does this implication hold?

Note that a proof for $\sigma(n^2)/q \mid n^2$ would imply the Descartes-Frenicle-Sorli Conjecture that $k=1$.

This is a partial answer:  I merely wanted to collect some more thoughts that recently occurred to me about this problem, after I have posted the question.

We have
$$\dfrac{N}{\sigma(n^2)/q} = \dfrac{q\sigma(q^k)}{2}$$
$$\dfrac{q^k n^2}{\sigma(n^2)/q} = \dfrac{q\sigma(q^k)}{2}$$
$$\dfrac{n^2}{\sigma(n^2)/q} = \dfrac{q}{2}\cdot{I(q^k)}.$$
Note that both $q/2$ and $1 < I(q^k) < 5/4$ are non-integers.
This does not necessarily mean, however, that
$$\dfrac{n^2}{\sigma(n^2)/q} = \dfrac{q}{2}\cdot{I(q^k)}$$
is a non-integer, as the RHS is equal to $(q+1)/2$ when $k=1$, which is an integer because $q \equiv 1 \pmod 4$.

I therefore conclude that a necessary and sufficient condition for $\sigma(n^2)/q \mid q^k n^2$ to imply $\sigma(n^2)/q \mid n^2$ is $k=1$.
