The unequal-GCD conjecture for odd perfect numbers

Let $N=q^k n^2$ be an odd perfect number in Eulerian form, where $q$ is the special prime and $\gcd(q,n)=1$. Consider the two greatest common divisors

$$\gcd(\sigma(q^k),\sigma(n^2))$$

and

$$\gcd(n^2,\sigma(n^2)).$$

Unequal-GCD conjecture. Unconditionally, these two quantities are unequal:

$$\gcd(\sigma(q^k),\sigma(n^2))\neq\gcd(n^2,\sigma(n^2)).$$

This is posed as a future research problem and remains open.