Gibbs free energy change at equilibrium
I'm really not getting why that post didn't answer OP's query.

Nevertheless, I'm posting the following argument as an answer.




  Why is the Gibbs free energy change equal to zero at equilibrium?


Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs 
Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.

Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,

$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$

where $\Delta n$ is the change in composition of the system.

Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.

$$\Delta_r G= 0\;.$$

Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$

The definition of equilibrium position sheds light on this:

A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .

This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$