It is easy to prove, using the relation $\prod_{d\mid n}d=n^{\sigma_0(n)/2}$ holds for $n\geq 1$ where $\sigma_0(n)$ is the number of divisors, the following
Proposition. The integer $n\geq 1$ is a perfect number if and only if $$2^{\sigma_0(n)}\cdot\left(\prod_{d\mid n}d\right)^2=\left(\sum_{d\mid n} d\right)^{\sigma_0(n)}$$
Thus it is easy to prove, noting the sum of divisors as $\sigma(n)$, that if $n\geq 1$ is an odd perfect number then $$\sigma\left(\sigma(n)^{\sigma_0(n)}\right)=\left(2^{\sigma_0(n)+1}-1\right)\cdot\sigma\left(\prod_{d\mid n}d^2\right)$$
My attempt was prove the converse statement,
Question. Prove that if an integer $n\geq 1$ satisfies $$\sigma\left(\sigma(n)^{\sigma_0(n)}\right)=\left(2^{\sigma_0(n)+1}-1\right)\cdot\sigma\left(\prod_{d\mid n}d^2\right)$$ then $n$ is odd and a perfect number (this is, an odd perfect number), or give a counterexample.
Thanks in advance.
This is only a partial answer to your question.
Suppose $n\geq 1$ satisfies $$\sigma\left(\sigma(n)^{\sigma_0(n)}\right)=\left(2^{\sigma_0(n)+1}-1\right)\cdot\sigma\left(\prod_{d\mid n}d^2\right).$$ Since the RHS is odd, then $$\sigma(n)^{\sigma_0(n)}$$ is a square or twice a square. (I think it is possible to rule out $$2m^2 = \sigma(n)^{\sigma_0(n)},$$ but I do not have enough time for this right now. I will come back to this particular case later.)
If $\sigma_0(n)$ is even, then $n$ is not a square.
Suppose that $$n = \prod_{i=1}^{\omega(n)}{{p_i}^{\alpha_i}}.$$ Then $$\sigma_0(n) = \prod_{i=1}^{\omega(n)}{\left(\alpha_i + 1\right)}.$$ Since $\sigma_0(n)$ is even and $n$ is not a square, then at least one $\alpha_i$ is odd. (I will stop here for now, I will come back to this question (and answer) later.)