Is it possible to construct a matrix : $A^2 = J_{n}$.

Note that $spectrum(J_n)=\{0,\cdot,0,n\}$; consequently, $spectrum(A)=\{0,\cdots,0,\pm\sqrt{n}\}$ and, necessarily, $trace(A)=\sqrt{n}$. Thus, $A$ may exist only when $n$ is a square. In this last case, $A$ exists

For $n=1$, $A=[1]$.

For $n=4$,

For $n=9$,

and so on...

If $n$ is a square (such as 4 in the example below), then the following pattern seems to work $$ \begin{bmatrix} 1& 1& 0& 0 \\ 0& 0 & 1&1 \\ 1&1 &0 &0 \\ 0&0 &1 &1 \end{bmatrix} $$ + many row-column permutations.

The answer is negative when $ n = 2 $. There, one finds by direct computation that the only matrices $ A $ which satisfy $ A^2 = J_2 $ are $ \pm \cfrac{\sqrt{2}}{2} J_2 $, which does not belong to the desired space.