Existence of a matrix whose power is $I$.

1. Let $m,n \ge 2$ be integers. Does there exist a matrix $A \in \mathbb{R}^{n\times n}$ such that $$A^m=I \ne A^k \ \forall\ k \in\{1,\ldots, m-1\}?$$ For $n=2,3,4$ and $m=3$ the answer is yes (see, e.g. Is there any matrix $2\times 2$ such that $A\neq I$ but $A^3=I$ and Find a $4\times 4$ matrix $A$ where $A\neq I$ and $A^2 \neq I$, but $A^3 = I$. and the various answers to those question)
2. Does there exist a matrix $A \in \mathbb{R}^{2\times2}$ which is not a rotation and such that $$A^3=I\ne A, A^2?$$
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If by "not a rotation" you mean $\vert\det\vert\ne 1$ then no. Powers of the matrix give powers of the determinant, and thus no possibility of $I$ in the power unless $\vert \det \vert = 1$ – adam W Jan 24 '13 at 1:42
I really meant a rotation, i.e. $a_{11}=a_{22}=\cos\theta, a_{12}=-a_{21}=-\sin\theta$ for some $\theta \in [0,2\pi)$. For $|\lambda|\ne 0,1$, the matrix $\text{diag}(\lambda,\lambda^{-1})$ is not a rotation, although it has determinant 1. – albmiz-mth Feb 3 '13 at 15:00
I thought that you may be thinking about that, so consider then the individual eigenvalues. They follow the same pattern in the powers, $\lambda \rightarrow \lambda^2 \rightarrow \lambda^3 \ldots$ and still no possibility of $I$ which has all eigenvalues of $1$. Unless they all have magnitude of $1$. Which would mean it is a rotation. – adam W Feb 3 '13 at 15:14

The $2 \times 2$ matrix representations of the complex $m$th roots of unity satisfy your definition, e.g.: $$z = e^{2\pi i/m}, \quad A = \left[ \begin{matrix} Re(z) & Im(z) \\ -Im(z) & Re(z) \end{matrix} \right]$$ Of course, you can trivially extend this to any size matrix by embedding it as a submatrix.
For arbitrary $n \in \{1, 2, 3, \ldots\}$ and $m = 2$, the rotation matrix works. For $m > 2$, just put a $2$-by-$2$ rotation matrix as a submatrix. (The answer is actually in the links you posted.)