Preamble: This post is an offshoot of this earlier MSE question.
The topic of odd perfect numbers likely needs no introduction.
Let $\sigma=\sigma_{1}$ denote the classical sum of divisors. Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$. A positive integer $x$ satisfying $I(x) < 2$ (respectively, $I(x) > 2$) is said to be deficient (respectively, abundant).
An odd perfect number $N$ is said to be given in Eulerian form if $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
The question is as is in the title:
If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(n^2)$ be abundant?
UPDATE (September 11, 2021 - 1:56 PM Manila time) - Finally proved that $\sigma(n^2)$ must be deficient in my answer below.
MY ATTEMPT
From the trivial relationship $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2,$$ I could only prove that $\sigma(n^2)/q^k \mid n^2 \mid N$ (since $\gcd(q^k, \sigma(q^k))=1$). This means that $\sigma(n^2)/q^k$ is deficient.
Also, I obtain $$\sigma(\sigma(q^k)\sigma(n^2)) = \sigma(2q^k n^2) = 3\sigma(q^k)\sigma(n^2).$$ Using the inequality $$\sigma(AB) \leq \sigma(A)\sigma(B)$$ which holds for all positive integers $A, B$, we have $$3\sigma(q^k)\sigma(n^2) = \sigma(\sigma(q^k)\sigma(n^2)) \leq \sigma(\sigma(q^k))\sigma(\sigma(n^2)).$$
From the paper titled On Odd Perfect Numbers and Even $3$-Perfect Numbers by Cohen and Sorli, published in Integers (2012, Theorem 3, page 3), we have the following result:
Let $q^k n^2$ be an odd perfect number with special prime $q$. Then the following inequality holds unconditionally: $$\sigma(\sigma(q^k)) \leq 3q^k - 1.$$
This result implies that $$3\bigg(\frac{q^{k+1} - 1}{q - 1}\bigg)\sigma(n^2) = 3\sigma(q^k)\sigma(n^2) = \sigma(\sigma(q^k)\sigma(n^2)) \leq \sigma(\sigma(q^k))\sigma(\sigma(n^2)) \leq (3q^k - 1)\sigma(\sigma(n^2)),$$ from which we get $$I(\sigma(n^2)) \geq 3\bigg(\frac{q^{k+1} - 1}{(q - 1)(3q^k - 1)}\bigg) = \frac{q - 3}{(q - 1)(3q^k - 1)} + \frac{q}{q - 1} := f(q,k).$$
I noticed that $$f(5,1) = \frac{9}{7}$$ $$f(13,1) = \frac{21}{19}.$$
I did confirm (graphically, by using WolframAlpha) that it is true that $$f(q,k) > 1.$$
My hunch is that $\sigma(n^2)$ is abundant, as $\omega(n) \geq 9$ (which means $n^2$ has a lot of prime factors), as proved by Nielsen (2015), where $\omega(n)$ is the number of distinct prime factors of $n$.
Alas, this is where I get stuck!
UPDATE
Since $3q^k - 1 < 3q^k$, then $$I(\sigma(n^2)) \geq f(q,k) = 3\bigg(\frac{q^{k+1} - 1}{(q - 1)(3q^k - 1)}\bigg) > 3\bigg(\frac{q^{k+1} - 1}{{3q^k}(q - 1)}\bigg) = I(q^k) > 1,$$ which analytically confirms my graphical observation above.
Let $N$ be an odd perfect number given in the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
From the trivial relationship $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2$$ we obtain $$\bigg(\frac{\sigma(q^k)}{2}\bigg)\cdot{\sigma(n^2)}=N=q^k n^2$$ which implies that $\sigma(n^2)$ is a proper factor of the (odd) perfect number $N$ (since $\sigma(q^k)/2 \geq (q^k + 1)/2 \geq (5+1)/2 = 3$).
We therefore conclude that $\sigma(n^2)$ must be deficient.