Now, I'm wondering how to prove this one. It seems simple but I might be looking in the wrong direction.
Properties which are used:
$\text{adj}(AB) = \text{adj}(B)\text{adj}(A)\:\ (1)$
$\text{adj}(I) = I\:\ (2)$
$X \implies Y$
Assuming $A$ is regular.
$AX = XA = I$, where $X$ is its inverse.
$\text{adj}(AX) = \text{adj}(XA) = \text{adj}(I)$
$\text{adj}(X)\text{adj}(A) = \text{adj}(A)\text{adj}(X) = I$, here the implication ends.
$\text{adj}(A)$ is regular and $\text{adj}(X)$ is its inverse.
$Y \implies X$
Assuming $\text{adj}(A)$ is regular.
$\text{adj}(A)X = X\text{adj}(A) = I$, where $X$ is the inverse ($X = \text{adj}(B)$).
According to $(1)$ and $(2)$, I can write this as...
$\text{adj}(BA) = \text{adj}(AB) = \text{adj}(I)$...
Now, here I'm stuck. I do not know if adjugate matrix is unique for each matrix. Don't know if I can just "inverse" the process to get the $BA = AB = I$, and show that $A$ is regular.