We denote the abundancy index of $x$ as
$$I(x) = \frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the sum of the divisors of $x$.
First, we prove the following claim:
Proposition 1.
$$I(n^2) \neq \frac{9}{5}$$
Proof.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and assume further that $I(n^2) = 9/5$.
This implies that $I(q^k) = 2/I(n^2) = 10/9$. Since $9 < 10 < \sigma(9) = 13$ and $\gcd(10, 9) = 1$, then the fraction $10/9$ is an abundancy outlaw. This contradicts $I(q^k) = 10/9$.
QED.
By Proposition 1, either $I(n^2) < 9/5$ or $I(n^2) > 9/5$ is true.
If $I(n^2) < 9/5$, we have the following result.
Proposition 2.
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5$$
Proof.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.
First, we show that $I(n^2) < 9/5 \Longrightarrow q = 5$:
$$\frac{2}{I(q^k)} = I(n^2) < \frac{9}{5} \Longrightarrow \frac{10}{9} < I(q^k) < \frac{q}{q - 1} \Longrightarrow q < 10.$$
That $q = 5$ follows from the fact that $q$ is the Euler prime (i.e., $q \equiv 1 \pmod 4$).
Next, we show that $I(n^2) > 9/5 \Longrightarrow q \geq 13$:
$$\frac{9}{5} < I(n^2) = \frac{2}{I(q^k)} \Longrightarrow 1 + \frac{1}{q} \leq I(q^k) < \frac{10}{9} \Longrightarrow q > 9.$$
That $q \geq 13$ follows from the fact that $q$ is the Euler prime.
In particular, we have shown that
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5.$$
QED.
The following result appears as Lemma 12 in "The third largest prime divisor of an odd perfect number exceeds one hundred" by Iannucci.
Proposition 3.
$$q = 5 \Longrightarrow k = 1$$
(This post is currently a WORK IN PROGRESS.)
