Question: > 
> ***Q***. Are there $n$-th root analogs of this non-diagonal cube-root of the $3 \times 3$
> identity matrix?
> 
> 
> 

\begin{align\*}
\left(
\begin{array}{ccc}
 0 & 0 & -i \\
 i & 0 & 0 \\
 0 & 1 & 0 \\
\end{array}
\right)^3
= \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)
\end{align\*}

I am looking for $A^n=I$, where $I$ is any dimension $\le n$.

(A naive question: I am not an expert in this area.)

Answer: 
We want to solve $A^n = I $ where $A$ and $I$ are $m \times m$

Let $ A = P \Lambda P^{-1} $

where $\Lambda = \text{diag}(\omega\_1, \omega\_2, \dots , \omega\_m) $

where the $\omega\_k$'s are $n$-roots of unity. That is, $\omega\_k = \exp\big(\dfrac{i 2 \pi j}{n} \big)$ ,$ k = 1,2,\dots,m $ and $ j \in \{ 0, 1, 2, ...., n-1 \} $

and $P$ is any invertible $ m \times m $ matrix.

Then $A^n = \big(P \Lambda P^{-1}\big)\big(P \Lambda P^{-1}\big)\dots \big(P \Lambda P^{-1}\big) = P^{-1} \Lambda^n P = P^{-1} I P = I $
The most general solution of the equation $A^n=1$ among $m\times m$ complex matrices is
$$
A=SDS^{-1},
$$
where $S$ is an invertible matrix and $D$ is a diagonal matrix whose diagonal entries are $n^{th}$ roots of unity.

The reason is that the minimal polynomial of such a matrix $A$ divides the polynomial $x^n-1$, and hence has distinct roots, which in turn implies that $A$ is diagonalizable.