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 $n = \gcd(n^2,\sigma(n^2))$.  (This follows from an unconditional proof for the condition
$$n = \gcd(n,\sigma(n^2)).$$
See below.)

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**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 in general.

Based from the results in this [MSE question](https://math.stackexchange.com/q/4341573/28816) 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 $I = n^j$ for some integer $j \geq 1$.

However, we know from the results of this [preprint](https://arxiv.org/abs/2202.08116) that
$$I = \frac{n}{\sigma(q^k)/2}\cdot\gcd(\sigma(q^k)/2,n).$$

We therefore have
$$n^{j-1} = \frac{\gcd(\sigma(q^k)/2,n)}{\sigma(q^k)/2}.$$

Since $j \geq 1$ 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 $j=1$.

Consequently, $\gcd(n,\sigma(n^2))=n$, which is equivalent to $\sigma(q^k)/2 \mid n$.