If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$? Or, in other words, if $A A^{-1} = I$ is satisfied, is there a possibility that $A^{-1} A \neq I$?
My suspicion is the answer is no.
 A: The result is not true for non-square matrices, since their left and right inverses are not equal.
A:  Short Answer 
Yes AA-1 = A-1A = I when the Det(A)$\neq$0 and A is a square matrix.
 Long Answer 
 A matrix is basically a linear transformation applied to some space. For the sake of simplicity I will assume that we are in a 2D plane having 2 basis vectors i^ and j^ each having the magnitude of 1 with coordinates (1,0) and (0,1) respectively. The columns of the matrix tells where the corresponding basis vectors lands after the transformation.
A-1A means that first we apply A transformation then we apply A-1 transformation. When we apply A transformation we reach some plane having some different basis vectors but after apply A-1 we again reach to the plane have basis i^ (0,1) and j^ (1,0). It means that after applying A-1 we reach to the transformation which does nothing. The identity transformation $
    \begin{bmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{bmatrix}$.
It is basically the plane we started before applying A transformation. That's why A-1A = I.
This is same as AA-1. It means that we first apply the A-1 transformation which will take as to some plane having different basis vectors. If we think what is the inverse of A-1 ? We are basically asking that what transformation is required to get back to the Identity transformation whose basis vectors are i^(1,0) and j^(0,1). Transformation A will do the job, since in this case after applying the A transformation we will reach to the identity transformation.
That's why 
AA-1 = I
