(Preamble #1: In what follows, we take $\sigma=\sigma_{1}$ to be the sum of the divisors, and denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.)
(Preamble #2: My sincerest apologies for the somewhat very long post -- I just had to put in all the details into one place for ease of quick reference later.)
PROBLEM
If $q^k n^2$ is an odd perfect number with Euler prime $q$, which of the following relationships between $q^2$ and $n$ hold?
(A) $$q^2 < \sigma(q^2) < n < \sigma(n)$$ (B) $$q^2 < n < \sigma(q^2) < \sigma(n)$$ (C) $$n < \sigma(n) < q^2 < \sigma(q^2)$$ (D) $$n < q^2 < \sigma(n) < \sigma(q^2)$$
PRELIMINARIES
(0) Of course, first of all, note that $q^2 \neq n$ since $\gcd(q,n)=1$.
(1) Note that $$\frac{q^2}{n}+\frac{n}{q^2} < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2\cdot\bigg(\frac{q^2}{n}+\frac{n}{q^2}\bigg).$$ Consequently, we know that $$\frac{q^2}{n}+\frac{n}{q^2} \text{ is bounded from above } \iff \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \text{ is bounded from above.}$$ In general, since the function $f(z) := z + (1/z)$ is not bounded from above, this means that we do not expect $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ to be bounded from above.
(2) $n < q^2 \implies k = 1$ [Dris, 2012]
(3) $$1 < I(q^2) \leq \frac{q^2 + q + 1}{q^2} = 1+\frac{1}{q}+\bigg(\frac{1}{q}\bigg)^2 \leq 1+\frac{1}{5}+\bigg(\frac{1}{5}\bigg)^2 = \frac{25+5+1}{5} = \frac{31}{25}$$ since $q$ being the Euler prime implies that $q$ is prime with $q \equiv 1 \pmod 4$, therefore $q \geq 5$.
Now, $1 < I(n) < 2$ since $n > 1$ is deficient, $n$ being a proper factor of the (odd) perfect number $q^k n^2$.
Consequently, $$1 < I({q^2}n) = I(q^2)I(n) = \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} < 2\cdot\frac{31}{25} = \frac{62}{25}.$$
(4) It follows from (1) and (3) that the following hold:
(a) $\sigma(n) \neq \sigma(q^2)$
Proof: Assume that $\sigma(n) = \sigma(q^2)$. Then we have (by (3)) $$I(q^2) = \frac{\sigma(q^2)}{q^2} = \frac{\sigma(n)}{q^2} < \frac{31}{25}$$ and $$I(n) = \frac{\sigma(n)}{n} = \frac{\sigma(q^2)}{n} < 2.$$ It follows that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2+\frac{31}{25}=\frac{81}{25}.$$ This implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).
(b) $\sigma(q^2) \neq n$.
Suppose to the contrary that $\sigma(q^2)=n$. Then we have (from (3)) $$1\cdot\frac{\sigma(n)}{q^2} = \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} < \frac{62}{25}.$$ It follows that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}=1+\frac{\sigma(n)}{q^2}<1+\frac{62}{25}=\frac{87}{25}.$$ This implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).
The proof for the following is similar to that of (b):
(c) $\sigma(n) \neq q^2$
MAIN RESULTS
The proofs of the succeeding three lemmas are trivial.
Lemma 1. The inequality $$I(q^2) + I(n) < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is true if and only if the biconditional $$q^2 < n \iff \sigma(q^2) < \sigma(n)$$ is true.
Lemma 2. The inequality $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < I(q^2) + I(n)$$ is true if and only if the biconditional $$q^2 < n \iff \sigma(n) < \sigma(q^2)$$ is true.
Lemma 3. The equation $$I(q^2) + I(n) = \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is true if and only if either $q^2 = n$ or $\sigma(q^2)=\sigma(n)$ hold.
We now prove our first main result:
Theorem 1. If $q^k n^2$ is an odd perfect number, then the inequality $$I(q^2) + I(n) < \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ holds.
Proof: Suppose to the contrary that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \leq I(q^2)+I(n).$$
Since $I(q^2) \leq 31/25$ and $I(n) < 2$, this implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2 + \frac{31}{25} = \frac{81}{25}$$ so that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).
By using Lemma 1, we are able to obtain the following Corollary to Theorem 1.
Corollary 2. If $q^k n^2$ is an odd perfect number, then the biconditionals $$q^2 < n \iff \sigma(q^2) < \sigma(n) \iff \frac{\sigma(q^2)}{n} < \frac{\sigma(n)}{q^2}$$ hold.
Proof: Trivial.
Note that Corollary 2 proves that the list of inequalities given in the Problem Statement exhausts all possible cases.
We now claim that:
Theorem 3. If $q^k n^2$ is an odd perfect number, then $q^2 < \sigma(n)$ and $n < \sigma(q^2)$ cannot be both true.
Proof:
Suppose that both $$1 < \frac{\sigma(q^2)}{n}$$ and $$1 < \frac{\sigma(n)}{q^2}$$ hold. Then it follows that both $$\frac{\sigma(n)}{q^2} < \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} = I({q^2}n) < \frac{62}{25}$$ and $$\frac{\sigma(q^2)}{n} < \frac{\sigma(q^2)}{n}\cdot\frac{\sigma(n)}{q^2} = I({q^2}n) < \frac{62}{25}$$ hold. This means that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} < 2\cdot\frac{62}{25} = \frac{124}{25}$$ which implies that $$\frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2}$$ is bounded from above, contradicting (1).
Note that Theorem 3 immediately rules out Case (B) and Case (D) above.
INQUIRIES
We are left to consider the following two (2) remaining cases:
(A) $$q^2 < \sigma(q^2) < n < \sigma(n)$$ (C) $$n < \sigma(n) < q^2 < \sigma(q^2)$$
Question #1 By this answer, we know (?) that the implication $n < q^{k+1} \implies k \neq 1$ is true if and only if $q^2 < n$. Can this result be improved?
Question #2 Following this answer, it is conjectured that $k \neq 1$. This would follow if we could rule out $q = q^k < \sigma(q) = \sigma(q^k) < n < \sigma(n)$. (That is, if we could prove that $\sigma(q^k) < n$ is false.) To what extent can a proof along this thread of thought be pursued, given Brown's arguments for a partial proof of $q^k < n$ in A Partial Proof of a Conjecture of Dris and the considerations in this MSE post?
