I think the density does go to zero, but quite slowly. If $p \equiv 1 \bmod 6$ is prime then there are two solutions $0<r<s<p-1$ of $$x^2+x+1=0 \bmod p$$
If $p\parallel n$ then, with probability $1,$ there are two distinct primes $x $ and $ y,$ each congruent to $r \bmod p,$ with $x \parallel n$ and $y \parallel n.$ ( Either or both could be congruent to $s$ as well.)
Then $p \parallel \gcd(n,\sigma(n^2))$ while $p^2 \parallel \gcd(n^2,\sigma(n^2)).$ So the asymptotic density for this not to happen is $1-\frac{p-1}{p^2}<1-\frac{1}{p-1}.$
If we can argue that the chance that none of these events happen is asymptotically $\prod(1-\frac{p-1}{p^2})$ over the primes congruent to $1 \bmod 6,$ then that asymptotic density is $0.$