Let $N$ and $M$ be two digit numbers. Then the digits of $M^2$ are those of $N^2$, but reversed. 
Let $N$ be a two digit number and let $M$ be the number formed from $M$ by reversing $N$'s digits. The digits of $M^2$ are precisely those of $N^2$, but reversed.

$Proof$:

Since $N$ is a two digit number, we can write $N = 10a + b$ where $a$ and $b$ are the digits of $N$. Since $M$ is formed from $N$ by reversing digits, $M = 10b + a$.
$N^2 = (10a + b)^2 = 100a^2 + 20ab + b^2 $. The digits of $N^2$ are $a^2, 2ab, b^2$.
$M^2 = (10b + a)^2 = 100b^2 + 20ab + a^2$. The digits of $M^2$ are $b^2, 2ab, a^2$, exactly the reverse of $N^2$.

This proposition is false. Let $N$ be $15$. That means the proof above is not correct, but I can't see where exactly.

As Greg's examples and other comments point out this, can only be true if $a^2,2ab$ and $b^2$ are all less than $10$. Otherwise there is a carryover that spoils it, as your example shows...
