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

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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
$$1 = \gcd(q^k n, 1) = \dfrac{\gcd\Bigg(n\sigma(n^2),i(q)\Bigg)}{i(q)} = \dfrac{\gcd\Bigg(n\sigma(n^2),\gcd(n^2,\sigma(n^2))\Bigg)}{i(q)} = \dfrac{\gcd\Bigg(\gcd(n\sigma(n^2),n^2),\sigma(n^2)\Bigg)}{i(q)} = \dfrac{\gcd\Bigg(n\gcd(\sigma(n^2),n),\sigma(n^2)\Bigg)}{i(q)}.$$

We consider two cases:

**Case 1:** $\gcd(n,\sigma(n^2))=1$

If $\gcd(n,\sigma(n^2))=1$, since it is known that $n^2 \nmid \sigma(n^2)$ and $\sigma(n^2) \nmid n$, so that we have
$$1 = \dfrac{\gcd\Bigg(n\gcd(\sigma(n^2),n),\sigma(n^2)\Bigg)}{i(q)}.$$
But we know that
$$3 \leq \gcd(n^2,\sigma(n^2)) = \gcd(n,\sigma(n^2)) = 1,$$
which is a contradiction.

**Case 2:** $\gcd(n,\sigma(n^2))=n$

So suppose to the contrary that $$I:=\gcd(n,\sigma(n^2))$$
and that
$$\gcd(n,\sigma(n^2))=n.$$
Since $\gcd(q^k n, 1) = 1$ holds in general, and because it implies that 
$$\gcd(nI,\sigma(n^2)) = \gcd(n^2,\sigma(n^2)),$$
then we have that
$$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2)) \implies \gcd(n,\sigma(n^2))=n.$$

This proves the claim.