Can we find $A^3 = (-A^t)^5$ for invertible $A\in \mathbb{R}^{4\times 4}$? I need to prove/disprove the existence of an invertible $4\times 4$ matrix over $\mathbb{R}$, $A\in \mathbb{R}^{4\times 4}$ such that
$$A^3 = (-A^t)^5$$
I think no such $A$ exists but was not able to disprove it without going over all Jordan forms and my guess is a simpler solution exists.
(Over $\mathbb{C}$ we have solutions and for $k\times k$ matrix with $k$ odd we can disprove by taking determinents)
 A: Let $Q$ be any $4\times4$ orthogonal matrix and $R$ be the $2\times2$ rotation matrix for the angle $\frac{\pi}{8}$. It is straightforward to verify that $A=Q(R\oplus R)Q^T$ indeed solves $A^3=-(A^T)^5$. We shall prove that all invertible solutions to the equation are in this form.
Given an invertible solution $A$, let $B=A^T$. Then
$$
B^3=(A^3)^T=(-B^5)^T=-A^5=-A^2A^3=-A^2(-B^5)=A^2B^5.
$$
Therefore
$$
A^2B^2=I.\tag{1}
$$
It follows that $A^3=-B^5=-(B^2)^2B=-(A^{-2})^2B=-A^{-4}B$. That is,
$$
B=-A^7.\tag{2}
$$
Hence $B$ commutes with $A$ and from $(1)$ we obtain
$$
(AA^T)^2=(AB)^2=A^2B^2=I.
$$
Since positive definite matrices have unique positive definite square roots, we see that $AA^T=I$. Hence $A$ is an orthogonal matrix and $(2)$ implies that $A^8=-I$. Thus $A$ admits the aforementioned decomposition $Q(R\oplus R)Q^T$.
A: 
(Over C we have solutions and for k×k matrix with k odd we can disprove by taking determinents)

And that is the key. First note that the function $\phi(re^{\theta i} ) = r \begin{bmatrix}\cos(\theta)&& -\sin(\theta) \\ \sin(\theta)&&\cos(\theta)  \end{bmatrix}$ is a homomorphism from complex numbers to $2 \times 2$ real matrices. Thus, every $n \times n$ complex matrix can be represented by a $2n \times 2n$ real matrix by turning all the complex numbers into blocks. Also note that if a $n \times n$ solution $A_n$ and a $m \times m$ solution $A_m$ exist, then the direct sum of $A_n \oplus A_m$ is a $(n+m) \times(n+m)$ solution. It therefore follows that if we can find a $1 \times 1$ complex solution, then we can use it to get a solution for any $n \times n$ over $\mathbb C$ and any $2n \times 2n$ over $\mathbb R$.
The set of $1\times1$ matrices is of course isomorphic to the set of scalars. The last fact we need is that $\phi(\bar z) = \phi(z)^T$. From there, we get the equation $re^{3\theta i} = -re^{-5\theta i}$ (the complex conjugate of $e^{xi}$ is $e^{-xi}$, so combining this fact that transposition corresponds to the conjugate, $(A^T)^5$ gives a $-5$ in the exponent). We can drop the $r$, as $z$ clearly is modulus $1$. $-1 = e^{\pi i}$, so we get $e^{3\theta i} = e^{(\pi -5\theta) i}$. Dropping the exponentiation and the $i$, we get $3\theta = \pi+ -5\theta \rightarrow 8\theta = \pi \rightarrow \theta = \frac{\pi}8$.
