Let $p^s Q^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv s \equiv 1 \pmod 4$ and $\gcd(p,Q)=1$.

I did some more digging on when the equations

$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$

simultaneously hold. Note that we have the identity

$$\gcd(\sigma(Q^2), \sigma(p^s)) \gcd(Q^2, \sigma(Q^2)) = \left(\gcd(Q, \sigma(Q^2))\right)^2.$$

**Hence, when exactly one of the three equations above holds, then the other two equations follow.**

In particular, note that
$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$

is equivalent to
$$\frac{Q^2}{\sigma(p^s)/2} = \frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2}$$

which, in turn, is equivalent to

$$Q = \gcd(\sigma(p^s)/2, Q).$$

This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.

Furthermore, in particular, note that
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$

is equivalent to

$$\left(\frac{Q}{\sigma(p^s)/2)}\right)\cdot\gcd(\sigma(p^s)/2, Q) = \frac{Q^2}{\sigma(p^s)/2}$$

which, in turn, is equivalent to

$$\gcd(\sigma(p^s)/2, Q) = Q.$$

This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.

Lastly, in particular, note that
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$

is equivalent to
$$\frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2} = \left(\frac{Q}{\sigma(p^s)/2}\right)\cdot\gcd(\sigma(p^s)/2, Q)$$

which, in turn, is equivalent to

$$\gcd(\sigma(p^s)/2, Q) = Q.$$

This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.

Thus, if we set
$$G = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$H = \gcd(Q^2, \sigma(Q^2))$$
$$I = \gcd(Q, \sigma(Q^2))$$

then we get the biconditional

$$G = H = I \iff Q \mid \sigma(p^s)/2.$$

Of course, as a sanity check, when $\sigma(p^s) = 2Q$, then we obtain the conjunction

$$Q \mid \sigma(p^s)/2$$

and

$$\sigma(p^s)/2 \mid Q,$$

which by Conjunction Elimination yields

$$Q \mid \sigma(p^s)/2$$

and hence, that

$$G = H = I.$$