Existence of Solutions to an Equation Involving the Sum-of-Divisors Function [Reference Request] Let $\sigma(x) = \sigma_1(x)$ denote the sum of all the positive divisors of $x$.
If $n \in \mathbb{N}$ is odd and $\gcd(n, \sigma(n)) = 1$, then do there exist any solutions to the following equation?
$$2{n^2}\sigma(n) = \sigma({n^2})\sigma(\sigma(n))$$
In other words, does there exist such an odd $N = {n^2}\sigma(n)$ which is also a perfect number?
Here, I also list some (open?) problems mentioned by other researchers:
(1)  Suryanarayana:  Is it true that every odd perfect number is of the form $m\sigma(m)$ for some odd integer $m$; if so, is $\gcd(m, \sigma(m)) = 1$ necessarily?
(2)  M. V. Subbarao:  Does every odd perfect number $n$ (if such exist) have the representation $$n = \frac{1}{2}m\sigma(m)  \hspace{0.5in}  (*)$$
Another question:  Whenever $n$ given by $(*)$ is perfect, does it follow that $n$ is odd and $\gcd(m, \sigma(m)) = 1$?
I would appreciate it if anybody could point me to reference(s) to recent results on either Problem (1) or Problem (2).
 A: (Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)
(Edited on March 12, 2015)
I was actually trying to (initially) rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$.  Just today, I found out that it might be possible to rule out the related condition $\sigma(n) = q$.
That being said, if we plug in $\sigma(n) = q$ into
$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$
with the (implicit) constraint $n = m^2$ (and since $n < \sigma(n) = q$ implies Sorli's conjecture that $k = 1$), then we get:
$$1 < I(q) \le \frac{6}{5} < \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n) < I(n^2) < 2$$
where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

The lower bound $$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$$ was communicated by Ochem to Dris in an e-mail dated April 17, 2013.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)
Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:
$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$
But $\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$.  Thus:
$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m)$$
WolframAlpha gives the approximation 
$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}}  \approx 1.20183425383797497745284556023594.$$
(Note that we have the equation $I(q^k)I(n^2) = I(q)I(n^2) = I(q)I(m^4) = 2$.)
We therefore have
$$I(q) \le \frac{6}{5} < \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m).$$
Note that $\gcd(q,n)=\gcd(q,m)=1$, so that we have $q \neq m$.  I now conjecture that $q < m = \sqrt{n}$ would follow from $I(q) < I(m)$ (and some further related inequalities), so that a contradiction against $n < \sigma(n) = q$ will arise.
