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).