A matrix equation $Ax=0$ has infinite solutions, does $A^Tx = 0$ have infinite solutions? I'm wondering whether a system with a transpose of a matrix has the same type of solution that the original matrix system has. If an equation $Ax=0$ equation has a unique solution, would a system with $A$ transpose instead of $A$ also have a unique solution? And what about with no solution, and infinite solutions?

If you are dealing with square matrices, then the answer is yes. You are essentially asking if the matrix is invertible or not, and $A$ is invertible iff $\det(A)\ne0$ and $\det(A)=\det\!\left(A^T\right)$.
However,
$$
\begin{bmatrix}
3&1&2\\
2&0&1
\end{bmatrix}
\begin{bmatrix}1\\1\\-2
\end{bmatrix}
=\begin{bmatrix}0\\0
\end{bmatrix}
$$
and any real multiple of this solution is a solution. Yet
$$
\begin{bmatrix}
3&2\\
1&0\\
2&1
\end{bmatrix}
x
=\begin{bmatrix}0\\0\\0
\end{bmatrix}
$$
requires $x=\begin{bmatrix}0\\0\end{bmatrix}$.
