Is it true that for any square matrix $A$, if $A^4=\text{Id}$ then $A^2=-\text{Id}$ or $A^2=\text{Id}$? I have tried some $A^2$ is not $-\text{Id}$ or $\text{Id}$ like $\pmatrix{1 & 0\\0 & -1}$, but I cannot find the corresponding matrix of $A$ in these cases.
 A: If you want a counterexample formed by a real matrix, then you can check that
with
$$
A=\left(\begin{array}{rrrr}
0&1&0&0\\
-1&0&0&0\\
0&0&-1&0\\
0&0&0&-1\end{array}\right)
$$
you get $A^4=I_4$. The matrix $A^2$ has two non-zero $2\times2$ blocks along the diagonal. The other block is $I_2$, the other $-I_2$, so $A^2\neq\pm I_4$.
A: Permutation matrix to the rescue! We take the permutation $1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 1$ and set entry $$a_{ij} = \begin{cases}1, &i\mapsto j\ \\ 0, &\text{otherwise}\end{cases}$$ to get the matrix
$$A = \begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\\\end{pmatrix}.$$
You can verify on WolframAlpha.
A: This in general is false because if
$$
A = \left(\begin{array}{cc}i&0\\0&1\end{array}\right)
$$
where $i$ is the square root of $-1$,
then
$A^4=I$, but
$$
A^2 = \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)
$$
this works because $i$ is one of the $4$ fourth roots of $1$, but it is not a square roots of $1$.
A: Like  @Giovanni Resta: 's answer 
$$
A = \left(\begin{array}{cc}0&-1&0\\1&0&0\\0&0&1\end{array}\right)
$$
