The short answer is NO, there are no more possible conditions under which $$\gcd(n,\sigma(n^2)) = \gcd(n^2,\sigma(n^2)),$$ other than $n = \gcd(n^2,\sigma(n^2))$.
Proof:
Consider $\gcd(q^k n, 1)=1$. This equation trivially holds.
Let $$i(q):=\gcd(n^2,\sigma(n^2)) = \dfrac{\sigma(n^2)}{q^k}.$$
We obtain $$1 = \gcd(q^k n, 1) = \dfrac{\gcd\bigl(n\sigma(n^2),i(q)\bigr)}{i(q)} = \dfrac{\gcd\bigl(n\sigma(n^2),\gcd(n^2,\sigma(n^2))\bigr)}{i(q)} = \dfrac{\gcd\bigl(\gcd(n\sigma(n^2),n^2),\sigma(n^2)\bigr)}{i(q)} = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$
We consider two cases:
Case 1: $\gcd(n,\sigma(n^2))=1$
If $\gcd(n,\sigma(n^2))=1$, then since it is known that $n^2 \nmid \sigma(n^2)$ and $\sigma(n^2) \nmid n$, so that we have $$1 = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$ But we know that $$3 \leq \gcd(n^2,\sigma(n^2)) = \gcd(n,\sigma(n^2)) = 1,$$ which is a contradiction.
Case 2: $\gcd(n,\sigma(n^2))=n$
We now prove that, indeed, $\gcd(n,\sigma(n^2)) = n$ holds.
Based from the results in this MSE question titled "On odd perfect numbers and a GCD - Part V", we have the following proposition:
THEOREM: If $m = q^k n^2$ is an odd perfect number with special prime $q$, then $$\gcd(n^2,\sigma(n^2))=\gcd(n^i,\sigma(n^2))$$ holds, where $i \geq 2$ is an integer.
From Case 1 above, we have the equation $$1 = \dfrac{\gcd\bigl(n\gcd(\sigma(n^2),n),\sigma(n^2)\bigr)}{i(q)}.$$
This means that $$\gcd(I\cdot{n},\sigma(n^2)) = \gcd(n^2,\sigma(n^2)),$$ where $I = \gcd(n,\sigma(n^2))$.
By the THEOREM above, we know that $I = n^i$ for some integer $i \geq 2$.
However, we know from the results of this preprint that $$I = \frac{n}{\sigma(q^k)/2}\cdot\gcd(\sigma(q^k)/2,n).$$
We therefore have $$n^{i-1} = \frac{\gcd(\sigma(q^k)/2,n)}{\sigma(q^k)/2}.$$
Since $i \geq 2$ is an integer, then the RHS is also an integer, so that $$\sigma(q^k)/2 \mid \gcd(\sigma(q^k)/2,n).$$
But we know that $$\gcd(\sigma(q^k)/2,n) \mid \sigma(q^k)/2$$ by the definition of GCD.
Hence, we do in fact have $$\gcd(\sigma(q^k)/2,n) = \sigma(q^k)/2$$ from which it follows that $i=1$.
Consequently, $\gcd(n,\sigma(n^2))=n$, which is equivalent to $\sigma(q^k)/2 \mid n$.