Find all $n\times n$ matrices with real entries such that $A^2=-I_n$.
If $A$ is such a matrix, since $(\det A)^2 = (-1)^n$, $n$ must be even.
Furthermore, $A$ annihilates $X^2+1 = (X-i)(X+i)$, so $A$ is diagonalizable over $\mathbb C$, with eigenvalues in $\{-i,i\}$. Let us write $A=PDP^{-1}$ where $D$ is a diagonal matrix with entries in $\{-i,i\}$ and $P$ is complex and non-singular.
Since the trace of $A$ is real, there must be the same number of $i$ and $-i$ on $D$'s diagonal.
Although interesting, this doesn't give a very explicit description of $A$.
I'm not even sure any product $$P\begin{pmatrix}
i\\
&\!\!i\\
&&\ddots\\
&&&-i\\
&&&&-i
\end{pmatrix}P^{-1}$$ yields a matrix with real entries, that's why I think something much more specific can be said about $A$.