Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it. $A =\left ( \begin{matrix}
    0 & 0 & 1 & 0\\
    1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
   \end{matrix} \right )$
Is there $B \in M_4({\mathbb{R}})$ such that $B^2$ is similar to $A$ ? If it is possible, find it.
The Jordan normal form of $A$ is $J=\begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0\\
   \end{pmatrix}$, $P=\begin{pmatrix}
    0 & 0 & 1 & 0\\
    0 & 1 & 0 & 0\\
    0 & 0 & 0 & 0\\
    1 & 0 & 0 & 1\\
   \end{pmatrix}$ such that $P^{-1}AP=J$
Then, $J=B^2$ but how can I find B? How do I know if it is possible to find $B$?

Your initial step of finding the Jordan form of $A$ is a helpful. With that, we can see that $A^3 = 0$, but $A^2 \neq 0$. This is enough for us to figure out that there is no $B$ such that $B^2$ is similar to $A$.
Indeed, if $B^2$ were similar to $A$, then it would follow that $B^6 = (B^2)^3$ similar to $A^3 = 0$, which means that $B$ is nilpotent, which means that the characteristic polynomial of $B$ must be $p(x) = x^4$. From this, it follows that $p(B) = B^4 = 0$. On the other hand, $B^4 = (B^2)^2$ must be similar to $A^2 \neq 0$. Thus, $B^4$ is similar to both the zero matrix and $A^2 \neq 0$, which means that $A^2$ is a non-zero matrix that is similar to the zero matrix, which is impossible.

The question of finding a general rule that determines whether $A$ has a square root is a bit trickier. As it turns out, this is true iff the blocks of the Jordan form of $A$ with length greater than $1$ can be paired off such that the sizes of the blocks in each pair are at most one apart.
