Here is an unconditional proof that $$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the OP, we have $$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

This is equivalent to $$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have $$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n^2, \sigma(n^2)) = i(q).$$

Consequently, we obtain $$G = i(q).$$

In particular, we get $$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

Here is a conditional proof that $$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the OP, we have $$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

This is equivalent to $$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have $$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(\sigma(n^2), n).$$

Consequently, we obtain $$G = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$

In particular, we get $$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2))$$ if and only if $$\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2)) = i(q).$$