Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$.
It is known that $$i(q)=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2},$$ where $i(q)$ is the index of the odd perfect number $N$ at the prime $q$.
Consider the quantity $\gcd(n,\sigma(n^2))$. Trivially, $$\gcd(n,\sigma(n^2)) \mid \gcd(n^2,\sigma(n^2))$$ holds since $n \mid n^2$ is true.
Want to show: $\gcd(n,\sigma(n^2)) = \gcd(n^2,\sigma(n^2))$
MY PROOF ATTEMPT #1
First, we write $n$ as $$n = i(q)\cdot\frac{\sigma(q^k)}{2n}.$$ Next, we express $\sigma(n^2)$ as $$\sigma(n^2) = i(q)\cdot{q^k}.$$ Thus, the GCD of $n$ and $\sigma(n^2)$ is given by $$\gcd(n,\sigma(n^2))=\gcd\left(i(q)\cdot\frac{\sigma(q^k)}{2n},i(q)\cdot{q^k}\right)=i(q)\cdot\gcd\left(\frac{\sigma(q^k)}{2n},q^k\right)=i(q)\cdot{1}=i(q)=\gcd(n^2,\sigma(n^2)),$$ since, in general, $\gcd(\sigma(q^k),q^k) = 1$ holds.
QED
MY PROOF ATTEMPT #2
Similar to the argument in Proof Attempt #1, we obtain $$\gcd(n,\sigma(n^2))=i(q)\cdot\gcd\left(\frac{\sigma(q^k)}{2n},q^k\right).$$
Consequently, we have the divisibility condition $$i(q)=\gcd(n^2,\sigma(n^2)) \mid \gcd(n,\sigma(n^2)).$$
Together with the divisibility constraint $$\gcd(n,\sigma(n^2)) \mid \gcd(n^2,\sigma(n^2))$$ and because of the fact that $\gcd(n,\sigma(n^2))$ and $\gcd(n^2,\sigma(n^2))$ are both positive, then it follows that $$\gcd(n,\sigma(n^2)) = \gcd(n^2,\sigma(n^2)).$$
Here then is my question:
Are these valid proofs for the equation $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$, if $N = q^k n^2$ is an odd perfect number with special prime $q$? If not, how can they be mended so as to produce correct arguments?
I am writing out the answer at the invitation of O.P.
The equation you have written for $\gcd(n, \sigma(n^2))$ in proof attempt number 1 seems to assume that $\gcd(i(q),n)=1$. If $\gcd(i(q),n)>1$, then that means $i(q)$ and $n$ share a common factor greater than $1$ and hence the expression $i(q) \times \sigma(q^k)/{2n}$ can be reduced further. If this expression can be reduced further then your equation does not hold. However, if $\gcd(i(q),n)=1$ then the expression $i(q) \times \sigma(q^k)/{2n}$ cannot be reduced any further and hence your equation holds. Therefore your equation holds on the condition that $\gcd(i(q),n)=1$.
Postscript
The O.P. has proven in the comment section that $\gcd(i(q),n)>1$.