REVERSEE...

This is not a solution, but an explanation of the changes to the problem.

The sentence "The two digits are non-zero and distinct" were added. As such, previously, the answers of 10 and 11 could have been considered correct. Those who entered these answers before Sept 1 have been marked correct.

With the change, the correct answer is 12. Those who enetered 12 have been marked correct.

The given answer of 13 was never correct, under any of the interpretations. Those who entered 13, but not 10, 11 or 12, have been marked incorrect.

A solution to this problem would be to establish that the given manipulations results in $\textrm{Rev}(n^2)=x^2 ~\forall \textrm{ 2 digit numbers }n=\overline{ab}$ where $a,b \leq 3$ . Restriction of $a,b$ doesn't affect the answer since we are looking for the smallest 2-digit $n$ .

NOTE: $\textrm{Rev}(n)$ refers to digit reversing function (a made up name).

So, we take $n=\overline{ab}$ where $a,b$ are the ten's and one's digit $(a,b\leq 3)$ of $n$ respectively. So, $n=10a+b$

We also have, $x=\overline{ba}=10b+a$

$n^2=(10a+b)^2=100a^2+10\cdot 2ab + b^2 = \overline{(a^2)(2ab)(b^2)} \\ \implies \textrm{Rev}(n^2)= \overline{(b^2)(2ab)(a^2)}=100b^2+10\cdot 2ab + a^2 = (10b+a)^2 = x^2$

So, $\boxed{\textrm{Rev}(n^2)=x^2 ~\forall \textrm{ 2 digit numbers }n=\overline{ab}\textrm{ with } a,b\leq 3}$

Now, that we have established this, we will just answer the smallest 2-digit number which has non-zero distinct digits and the digits are $\leq 3$ . The answer trivially comes out to be $\boxed{12}$