If $N = q^k n^2$ is an odd perfect number with special prime $q$, then must $\sigma(q^k)$ be deficient? The topic of odd perfect numbers likely needs no introduction.
Let $\sigma=\sigma_{1}$ denote the classical sum of divisors.  Denote the abundancy index by $I(x)=\sigma(x)/x$.
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 $\sigma(q^k)$ be deficient?

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 $\sigma(q^k)/2 \mid n^2 \mid N$ (since $\gcd(q^k, \sigma(q^k))=1$).  This means that $\sigma(q^k)/2$ is deficient.
I could do
$$\sigma(q^k)=\dfrac{2n^2}{\dfrac{\sigma(n^2)}{q^k}},$$
and then iterate to obtain
$$\sigma(\sigma(q^k))=\sigma\Bigg(\dfrac{2n^2}{\dfrac{\sigma(n^2)}{q^k}}\Bigg).$$
Finally,
$$I(\sigma(q^k))=\dfrac{\sigma(\sigma(q^k))}{\sigma(q^k)}=\dfrac{\sigma\Bigg(\dfrac{2n^2}{\dfrac{\sigma(n^2)}{q^k}}\Bigg)}{\dfrac{2n^2}{\dfrac{\sigma(n^2)}{q^k}}},$$
and this is where I get stuck since I do not know how to simplify the complex fraction on the RHS, with the end goal of obtaining (hopefully) tight lower and upper bounds, for $I(\sigma(q^k))$.
I do know however, that the following relationship and estimate hold:
$$\dfrac{\sigma(n^2)}{q^k} \mid n^2$$
$$\dfrac{\sigma(n^2)}{q^k} \geq 3.$$
Note that, for the case of even perfect numbers $M = (2^p - 1)\cdot{2^{p-1}}$ (where $2^p - 1$, and therefore $p$, is prime),
$$I(\sigma(2^p - 1)) = I(2^p) = 2 - \dfrac{1}{2^p} \leq \frac{7}{4},$$
or by observing directly that $\sigma(2^p - 1) = 2^p$ is a prime power and therefore deficient.
Update (September 29, 2020) - As correctly observed by mathlove, the last inequality above should be
$$I(\sigma(2^p - 1)) = I(2^p) = 2 - \dfrac{1}{2^p} \geq \frac{7}{4}.$$
Assume that $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$ is a prime power.  This forces $\sigma(q^k)=2$.  This means $2\sigma(n^2)=2q^k n^2$, which in turn is equivalent to
$$I(n^2) = q^k < 2$$
which is a contradiction (as $q$ being the special prime it ought to satisfy $q \equiv k \equiv 1 \pmod 4$, which means $q \geq 5$ and $k \geq 1$, and these imply that $q^k \geq 5$).  Thus, $\sigma(q^k)$ is not a prime power.
Assume that $\sigma(q^k) = u^s v^t$, where $u < v$ are primes.  Since $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$, then this forces $u = 2$ and $s = 1$.  (Note that, WLOG, we may assume that $v \geq 3$.)  We now compute
$$I(\sigma(q^k))=\frac{3}{2}\cdot{I(v^t)}<\frac{3}{2}\cdot{\dfrac{v}{v-1}} \leq \bigg(\frac{3}{2}\bigg)^2 = \frac{9}{4},$$
whence we cannot conclude if $\sigma(q^k)$ is deficient or not.
Alas, this is where I get stuck.
 A: On OP's request, I am converting my comment into an answer.
Also, I'm going to add some more thoughts at the end of this answer.
I noticed the following :
(1) In the case of even perfect numbers, we have
$$\frac 74\color{red}{\le} I(\sigma(2^p−1))=2−\frac{1}{2^p}\lt 2$$
from which we see that $\sigma(2^p−1)$ is deficient.
(2) If $\sigma(q^k)=u^sv^t$ where $u\lt v$ are primes such that $5\color{red}{\le} v$, then $I(\sigma(q^k))<\dfrac 32\cdot \dfrac{v}{v−1}\lt 2$, so $\sigma(q^k)$ is deficient.
(3) If $\sigma(q^k)=u^s\cdot 3^t$ where $u$ is an integer (not necessarily a prime) such that $\gcd(u,3)=1$, then $\dfrac u2\ (=m)$ is odd with $s=1$, and $$I(\sigma(q^k))=\dfrac{3\sigma(m)(3^{t+1}-1)}{2m\cdot 3^t\cdot 2}=\underbrace{\dfrac 34\left(3−\dfrac{1}{3^t}\right)}_{\ge 2}\cdot \underbrace{\dfrac{\sigma(m)}{m}}_{\ge 1}\ge 2$$ so $\sigma(q^k)$ is not deficient.

In the following, I'm going to add some more thoughts.
(4) One can prove that if $(q,k)$ satisfies either $q\equiv 2\pmod 3$ or $(q,k)\equiv (1,2)\pmod 3$, then $\sigma(q^k)$ is not deficient.
Proof :
If $q\equiv 2\pmod 3$, then we have
$$\sigma(q^k)=1+q+\cdots +q^k\equiv (1-1)+(1-1)+\cdots +(1-1)\equiv 0\pmod 3$$since $k$ is odd.
Also, if $(q,k)\equiv (1,2)\pmod 3$, then we have
$$\sigma(q^k)=1+q+\cdots +q^k\equiv 1+1+\cdots +1\equiv k+1\equiv 0\pmod 3$$
So, in either case, we get $\sigma(q^k)\equiv 0\pmod 3$.
Since we have $\sigma(q^k)\equiv 2\pmod 4$, there are positive integers $s,t$ such that $$\sigma(q^k)=2s\cdot 3^t$$where $s$ is odd satisfying $\gcd(s,3)=1$. Then, we have
$$I(\sigma(q^k))=\frac{3\sigma(s)(3^{t+1}-1)}{2s\cdot 3^t\cdot 2}=\underbrace{\dfrac 34\left(3−\dfrac{1}{3^t}\right)}_{\ge 2}\cdot \underbrace{\dfrac{\sigma(s)}{s}}_{\ge 1}\ge 2$$
from which we see that $\sigma(q^k)$ is not deficient.

So, the remaining cases are $(q,k)$ satisfying either $(q,k)\equiv (1,0)\pmod 3$ or $(q,k)\equiv (1,1)\pmod 3$.
A: 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
$$I(\sigma(q^k)) \leq \dfrac{3}{I(q^k)} - \dfrac{1}{\sigma(q^k)},$$
from which it follows that
$$I(\sigma(q^k)) \leq \dfrac{3}{I(q^k)} - \dfrac{1}{\sigma(q^k)} < \dfrac{3}{I(q^k)}.$$
But we know that
$$I(q^k) \geq \dfrac{q+1}{q} \iff \dfrac{3}{I(q^k)} \leq \dfrac{3q}{q+1} < 3.$$
We therefore infer that $I(\sigma(q^k)) < 3$, from which we cannot conclude whether $\sigma(q^k)$ is deficient or not.
