(Pardon me for being somewhat stubborn, but this question will be the last for this week. This post is an offshoot of this one.)
Let $N = q^k n^2$ be an odd perfect number be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Set $$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H := \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ $$I := \gcd\bigg(n,\sigma(n^2)\bigg).$$
It is known (and fairly easy to prove) that $$GH = I^2.$$
It is also known (and also very easy to prove) that the divisibility chain $$G \mid I \mid H$$ holds.
Here is my:
INITIAL QUESTION: What is the value of $$\frac{I}{G}=\frac{H}{I}?$$
MY ATTEMPT
I claim that $$\dfrac{I}{G} = \dfrac{H}{I} = \dfrac{n}{\sigma(q^k)/2},$$ if and only if $\sigma(q^k)/2 \mid n$.
To this end, initially it must be the case that
$$\sigma(q^k)/2 \mid n \iff n \mid \sigma(n^2),$$
basically because of the equation
$$\frac{q^k n}{\sigma(q^k)/2} = \frac{\sigma(n^2)}{n}.$$
But we also have
$$n \mid \sigma(n^2) \iff G = \sigma(q^k)/2$$
from Theorem B in this post.
Since $H$ is also the index of $N$ at the prime $q$, then
$$H = \frac{n^2}{\sigma(q^k)/2}.$$
Also, we obtain
$$n \mid \sigma(n^2) \iff I = n.$$
We therefore obtain
$$\dfrac{I}{G} = \dfrac{H}{I} = \dfrac{n}{\sigma(q^k)/2},$$
which is an integer, if and only if $\sigma(q^k)/2 \mid n$.
FINAL QUESTION: What is the value of $$\frac{I}{G}=\frac{H}{I}$$ if $\sigma(q^k)/2$ does not divide $n$?
On OP's request, I am converting my comment into an answer.
FYI, one can at least say that $$\begin{align}\dfrac HI&=\dfrac{n^2}{\gcd\bigg(n,\sigma(n^2)\bigg)\sigma(q^k)/2}=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,\sigma(n^2)\sigma(q^k)/2\bigg)} \\\\&=\dfrac{n^2}{\gcd\bigg(n\sigma(q^k)/2,n^2q^k\bigg)}=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,nq^k\bigg)} \\\\&=\dfrac{n}{\gcd\bigg(\sigma(q^k)/2,n\bigg)}\end{align}$$