If $N = q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $q=5$.

Erratum (late update):  As pointed out by Joshua Zelinsky (as well as by Carl Pomerance), the argument presented in the answer in this post requires
$$x = \frac{2\varphi(n)}{n} > 1,$$
an inequality which is currently not known (as of March 25, 2020).

MSE QUESTION

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.  That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

From a comment underneath this earlier question, we have the equation (and corresponding inequalities)

$$1 < \frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1} \leq \frac{5}{4}$$

since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$.

This implies that

$$\frac{4}{5} \leq \frac{\frac{\varphi(N)}{N}}{\frac{\varphi(n)}{n}} = \frac{q-1}{q} < 1.$$

But from the following source:

Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds

$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$

However, we also have

$$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$

Notice that

$$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$  

Therefore, we have the bounds

$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$

and

$$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$

which implies that

$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$

WolframAlpha gives the rational approximation

$$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$

Here is my question:

Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$?

MOTIVATION FOR THE INQUIRY

It can be shown that the equation

$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$

together with the bounds

$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}$$

and

$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}$$

imply

$$0.92 \approx \frac{\frac{120}{217\zeta(3)}}{\frac{1}{2}} < \frac{q}{q-1} < \frac{\frac{5}{8}}{\frac{120}{217\zeta(3)}} \approx 1.358574729,$$

from which we obtain trivial bounds.

Nonetheless, it can be shown that the equation

$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$

together with the upper bound $\varphi(N)/N < 1/2$ implies that

$$q < \frac{x}{x-1}$$

where

$$x = \frac{2\varphi(n)}{n}.$$

Thus, if we can improve the upper bound for $\varphi(N)/N$ to something smaller than $1/2$ (say $1/2 - \varepsilon$ for some tiny $\varepsilon > 0$), then we can improve the coefficient of $\frac{\varphi(n)}{n}$ in $x$ to some number bigger than $2$.  Likewise, if we can get a better lower bound for $\varphi(N)/N$, then we will be able to get an improved lower bound for $\varphi(n)/n$.  Together, they would translate (hopefully!) to a numerical upper bound for the special/Euler prime $q$!

POSTED ANSWER

Eureka!!!

Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.

From the equation and lower bound for $\varphi(N)/N$

$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$

and the equation

$$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$

we get the lower bound

$$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$

This implies that we have the upper bound

$$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$

which can be solved using WolframAlpha, yielding the upper bound

$$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$

from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.  

QED

REVISED BOUNDS FOR THE ABUNDANCY INDICES OF DIVISORS OF ODD PERFECT NUMBERS

Since $q=5$ holds, then since $q \mid q^k$ (for all positive integers $k$), then

$$\frac{q+1}{q} = I(q) \leq I(q^k) < \frac{q}{q - 1} < \frac{2(q - 1)}{q} < \frac{2}{I(q^k)} = I(n^2) \leq \frac{2q}{q + 1},$$

where $I(x)=\sigma(x)/x$ is the abundancy index of the positive integer $x$ (and $\sigma(x)$ is the sum of divisors of $x$).  We therefore have the revised bounds

$$\frac{6}{5} \leq I(q^k) < \frac{5}{4} < \frac{8}{5} < I(n^2) \leq \frac{5}{3}.$$

Note that we then have

$$\bigg(I(q^k) - \frac{6}{5}\bigg)\bigg(I(n^2) - \frac{6}{5}\bigg) \geq 0$$

which implies

$$I(q^k)I(n^2) - \frac{6}{5}\bigg(I(q^k) + I(n^2)\bigg) + \bigg(\frac{6}{5}\bigg)^2 \geq 0$$

from which it follows that

$$\frac{43}{15} = \frac{5}{3} + \frac{6}{5} = 2\cdot\frac{5}{6} + \frac{6}{5} \geq I(q^k) + I(n^2).$$

We also have the lower bound

$$I(q^k) + I(n^2) > \frac{57}{20}.$$

Better values/bounds are known when the Descartes-Frenicle-Sorli Conjecture that $k=1$ is assumed true (or otherwise), given that $q=5$ holds.  (See the following MSE question for more information: On the Descartes-Frenicle-Sorli conjecture and the Euler prime of odd perfect numbers.)