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