If $A^TA=cI$ for some scalar $c\neq0$, then $A^T=cA^{-1}$, and thus $AA^T=cI$ as well. (The first equation says that $c^{-1}A^T$ is a left-inverse of $A$. And a left-inverse is also a right-inverse: If $AB = I$ then $BA = I$ .) So what if $c=0$?
Over the real numbers, $A^TA=0$ implies $A=0$, since $A^TA$ contains the sum of squares of elements of each column of $A$.
Over a field with characteristic $p$, the $p\times p$ matrix
$$A=\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&\vdots\\1&1&\cdots&1\end{bmatrix}$$
(for example) has $A^TA=pA=0$ even though $A\neq0$. But in this case $AA^T=0$ as well. Does that always happen?
No.
Suppose $i^2=-1$ for some $i$ in the field (e.g. $i=2$ in the finite field of size $5$), and consider
$$A=\begin{bmatrix}1&0\\i&0\end{bmatrix}.$$