If $q^k n^2$ is an odd perfect number with Euler prime $q$, is it possible to have $n\sigma(n)=q\sigma(q^k)$? QUESTION

If $q^k n^2$ is an odd perfect number with Euler prime $q$, is it possible to have $n\sigma(n)=q\sigma(q^k)$?

BACKGROUND
If $\sigma(N)=2N$ (where $\sigma(N)$ is the sum of the divisors of $N$), then $N$ is said to be perfect. (Denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.)
Euler proved that every odd perfect number has the form $N=q^k n^2$, where $q$ is prime (called the Euler prime) satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
MY ATTEMPT TO ANSWER THE QUESTION
Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$, and assume to the contrary that $n\sigma(n)=q\sigma(q^k)$.  Since $\gcd(q,n)=1$, this implies that
$$\frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n} \in \mathbb{N}.$$
Since $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$ and $n$ is odd, then $\sigma(q^k) \neq n$.  This implies that
$$2 \leq \frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n}.$$
But $qn$ and $q^k n$ are both deficient.  This means that
$$\frac{\sigma(q)}{n}\cdot{2} \leq \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} = I(qn) < 2 \implies \frac{\sigma(q)}{n} < 1 \implies q < n,$$
and
$$\frac{\sigma(n)}{q^k}\cdot{2} \leq \frac{\sigma(n)}{q^k}\cdot\frac{\sigma(q^k)}{n} = I({q^k}n) < 2 \implies \frac{\sigma(n)}{q^k} < 1 \implies n < q^k.$$
Together, $q < n$ and $n < q^k$ imply that $k > 1$.  Additionally, the equation
$$n\sigma(n) = q\sigma(q^k)$$
is equivalent to
$$\frac{\sigma(n)}{\sigma(q^k)}=\frac{q}{n},$$
which together with $q < n$, implies that
$$\sigma(n) < \sigma(q^k).$$
Consequently, we have
$$q < \sigma(q) < n < \sigma(n) < q^k < \sigma(q^k)$$
so that
$$\frac{\sigma(q)}{n} < 1 < \frac{\sigma(n)}{q}$$
and
$$\frac{\sigma(n)}{q^k} < 1 < \frac{\sigma(q^k)}{n},$$
whereupon I do not obtain any contradictions.
 A: This isn't a answer, only the calculations that I can get (by the nature of the problem concerning odd perfect numbers, one can do deductions with the purpose to prove a more elaborated statement or well with the purpose to finish a proof by contradiction of the statement) . I hope that if there are mistakes some user can tell me. 
I prefer the notation $N=q^{4\lambda+1}m^2$, where $q^{4\lambda+1}$ is the Euler's factor associated to the odd perfect number. Since the sum of divisors function $\sigma(n)$ is multiplicative and $N$ is perfect we've the factorization $$\sigma(q^{4\lambda+1})\sigma(m^2)=2q^{4\lambda+1}m^2.\tag{1}$$ On the other hand we've your condition 
$$m\sigma(m)=q\sigma(q^{4\lambda+1}).\tag{2}$$ Thus multiplying by $\sigma(m^2)$ your condition $(2)$ with the purpose to combine with $(1)$, we get from $$m\sigma(m)\sigma(m^2)=q\sigma(N)=2q^{4\lambda+2}m^2,$$ and thus this 

Claim 1. Under previous assumptions $$\sigma(m)\sigma(m^2)=2q^{4\lambda+2}m.\tag{3}$$

Now we search a more elaborated statement combining previous claim with the fact that $$\sigma(m)\cdot\sigma(m^2)\geq (1+m)\cdot(1+m^2+(\sigma(m)-1)),$$ to get $$m^2+\sigma(m)\leq 2q^{4\lambda+2}\frac{m}{m+1},$$ and thus $m^2+\sigma(m)< 2q^{4\lambda+2}$, since $m/(m+1)<1$. From this, we can write 

Claim 2. Under previous assumptions $$\sigma(m)<2q^{4\lambda+2}-m^2.$$

A: (What follows are just partial results for the present problem.)
Building from the results in the answer to this MSE question, I have
$$\frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n}=x \in \mathbb{N}.$$
If the immediately preceding equation is true, then
$$xn = \sigma(q^k)$$
$$\sigma(n) = xq$$
$$\sigma\left(\frac{1}{x}\sigma(q^k)\right)=xq.$$
There are no solutions to this last equation for $k>1$ and $k \equiv 1 \pmod 4$, for particular values of $x$.  We have $\sigma(X) \geq X$, where the inequality is strict for $X>1$;  hence,
$$xq=\sigma\left(\frac{1}{x}\sigma(q^k)\right) \geq \frac{1}{x}\sigma(q^k) \geq {\frac{1}{x}}{q^k},$$
so we would need
$$q^{k-1} \leq x^2.$$
But for $k>1$ and $k \equiv 1 \pmod 4$, we have $k \geq 5$, so that
$$q^{k-1} \geq q^4 \geq 5^4.$$
Hence the original system of equations will only have a solution when
$$x^2 \geq 5^4 \implies x \geq 25.$$
Consequently, we obtain
$$25 \leq \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}.$$
Multiplying both sides of the inequality and equation by $\sigma(n)/{q^k}$, we get
$${25}\cdot\frac{\sigma(n)}{q^k} \leq \frac{\sigma(n)}{q}\cdot\frac{\sigma(n)}{q^k} = \frac{\sigma(q^k)}{n}\cdot\frac{\sigma(n)}{q^k} = I({q^k}n) < 2.$$
Multiplying both sides of the inequality and equation by $\sigma(q)/n$, we get
$${25}\cdot\frac{\sigma(q)}{n} \leq \frac{\sigma(q)}{n}\cdot\frac{\sigma(q^k)}{n} = \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} = I(qn) < 2.$$
Therefore, we have $\sigma(n)/{q^k} < 2/{25}$ and $\sigma(q)/n < 2/{25}$, from which it follows that
$$\frac{\sigma(n)}{q^k} < \frac{2}{25} < 25 \leq \frac{\sigma(q^k)}{n}$$
and
$$\frac{\sigma(q)}{n} < \frac{2}{25} < 25 \leq \frac{\sigma(n)}{q},$$
whereupon I still do not get any contradictions.
A: This is only a partial answer and serves to add perspective to the problem.
Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.
Since it is not possible to have both
$$\frac{\sigma(n)}{q} < 1 < \frac{\sigma(q)}{n}$$
and
$$\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(n)}{q^k},$$
then we know that
$$\lnot\bigg(\left(\frac{\sigma(n)}{q} < 1 < \frac{\sigma(q)}{n}\right) \land \left(\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(n)}{q^k}\right)\bigg)$$
must be true.
Therefore, it must be the case that
$\bigg(\frac{\sigma(q)}{n} < 1\bigg) \lor \bigg(\frac{\sigma(n)}{q} \geq 1\bigg) \lor \bigg(\frac{\sigma(n)}{q^k} \leq 1\bigg) \lor \bigg(\frac{\sigma(q^k)}{n} \geq 1\bigg).$ (A)
Towards the end of the original post, it is shown that the problem at hand is equivalent to proving that not all disjuncts in (A) are true.
