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: 
You can take a rotation matrix that rotates $\phi=2\pi/n$ around some axis, for example in 3 dimensions:

$$R\_\phi=\begin{pmatrix}
\cos \phi & \sin\phi & 0 \\
-\sin\phi & \cos\phi &0 \\
0 & 0 & 1\\
\end{pmatrix}$$

This won't work for $n=2$ though, because $R\_\phi$ is diagonal in that case like $R\_\phi=\operatorname{diag}(-1,-1,1)$.

Then $R\_\phi^n=I$ and $R\_\phi^k\neq I$ for any $0<k<n$. You can build new matrices using any invertible matrix $A$ and conjugate like

$$R\_{\phi,A}:= AR\_\phi A^{-1} \tag 1$$

then obviously:

$$R\_{\phi,A}^n = (AR\_\phi A^{-1})^n = A R\_\phi^nA^{-1} = I$$

(I am not sure whether this could fix the case $n=2$ and produce some valid (non-diagonal) results.)
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.