Let $k>n>0$ be two integers. Let $A$ be a real $k \times k$ matrix. We suppose that $Dim(Ker(A))>0$. We look at the system
$$Ax=0$$
We know that there is infinitely many solutions to this system because of $Dim(Ker(A))>0$. But can we always find $k-n$ different solutions $x^{(1)}=(x_1^{(1)}, \ldots, x_{k}^{(1)}),\ldots, x^{(k-n)}=(x_1^{(k-n)}, \ldots, x_{k}^{(k-n)})$ satisfying $x_i^{(i)} \neq 0$ for $i=1,\ldots, k-n$? If not, what extra conditions should we add on $A$ to get this property ?
No. Suppose for example that $A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$. Then the kernel of $A$ is the linear span of $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. In particular, every $x \in \ker(A)$ satisfies $x_2 = 0$.
A sufficient (but of course not necessary) condition to achieve what you want would be that every combination of $k-1$ columns is linearly independent. In that case your eigenvectors don't have a zero entry and you get what you want trivially.
Also note that your property is not basis-independent. So almost every change of basis would achieve this as well.