Let's determine the distinct digits $A, B$ such that the following multiplication holds: $A \cdot \overline{A B} = \overline{B A A}$ (in the decimal system).
In the decimal system, $\overline{A B}=10 A+B$ and $\overline{B A A}=100 B+11 A$, according to the problem,

$$
A(10+B)=100 B+11 A
$$

or

$$
A(B-1)=10\left(10 B+A-A^{2}\right)
$$

Here, the right side is divisible by 10, so $A(B-1)$ must also be divisible by 10. Since $A$ and $B-1$ are less than 10, this can only happen if one of the factors is 0, or if one factor is 2 and the other is 5. $A$ cannot be 0, since $\overline{A B}$ is a genuine two-digit number, and if $B-1=0$, then from (2)

$$
A^{2}-A=A(A-1)=10
$$

would be true, which does not provide a solution, since 10 cannot be expressed as the product of two consecutive integers.

If $A=5$, then from (1) we get $\frac{39}{19}$ for $B$, which is not a solution to our problem since it is not an integer. If $B-1=5$, then from (1)

$$
A^{2}=60+\frac{A}{2}
$$

Here, the right side is between 60 and 65, so the only possible solution in the set of natural numbers is $A=8$, and by substitution, we can verify that this is indeed a solution with $B=6$: $8 \cdot 86=688$.

Therefore, the only solution to the problem is: $A=8, B=6$. (In our solution, we did not use the condition of the problem that the digits $A, B$ must be different.)