Find all natural numbers $n \geq 3 $ such that if matrix $A$ has certain properties (e.g. $A^n = I$) then $A = CBC^{-1}$ Find all natural numbers $n \geq 3 $ such that:
if: $A$ is a $2 \times 2$ matrix with real coefficients $\land \space A^n = I \land \space A^j \neq I$ for $ 0 <j <n$, 
then: there are such invertible matrices $B$ and $C$ that $A = CBC^{-1}$ and 
\begin{equation}
  B =
      \begin{bmatrix}\cos\space t & \sin \space t \\-\sin \space t & \cos \space t \end{bmatrix}
\end{equation}
for certain $t \in \Bbb R $ 
I'm not really sure how to approach such problem. I guess one is supposed to use the properties of the determinant and the fact that 
$detB = $cos$^2(t)+$sin$^2(t)=1 $
 A: Let us start by calculating $A^k$. By our identity $A=CBC^{-1}$, we can conclude that $$A^k=(CBC^{-1})\cdots(CBC^{-1})=CB(C^{-1}C)B(C^{-1}C)\cdots(C^{-1}C)BC^{-1}=CB^kC^{-1}.$$ If $A^n=CB^nC^{-1}=I$, we can bring $C,C^{-1}$ to the right site to deduce $B^n=1$. Moreover, if $A^k\neq I$, then $$I\neq A^k=CB^kC^{-1}\Longrightarrow I=C^{-1}C\neq C^{-1}A^kC=B.$$ To sum up, we may equivalently ask for the existence of $t\in \mathbb R$ such that $B=B(t)$ satisfies $B^n=I$, $B^k\neq I$ for $k<n$.
There are many ways to find such values of $t$, here is one that I particularly like: The matrix $B(t)$ corresponds to the complex number $\cos(t)+i\sin(t)=\exp(it)$ which has length one (basically by the formula you stated). So the question is to find an $n$-th root of unity in $\mathbb C$. These roots are well-known to be $$\zeta_n=\exp\left(\frac{2\pi i}{n}\right)=\cos\left(\frac{2\pi}{n}\right)+i\sin\left(\frac{2\pi}{n}\right).$$ Each $\zeta_n$ has the property that $\zeta_n^n=1$ but $\zeta_n^k\neq 1$ for $k<n$.
Consequently, if you let $t=\frac{2\pi}{n}$, your matrix $B$ does the job and provides an example for such a matrix $A$ by choosing $C=I$. Note that we even know $B^k$ in this case: Since $\zeta_n^k=\exp\left(\frac{2k\pi i}{n}\right),$ $$B^k=\begin{pmatrix}\cos\left(\frac{2k\pi}{n}\right)&\sin\left(\frac{2k\pi}{n}\right)\\ -\sin\left(\frac{2k\pi}{n}\right) &\cos\left(\frac{2k\pi}{n}\right)\end{pmatrix}.$$
Let me know if you have any questions.
EDIT I just saw that I had a slight misunderstanding of the question. What I showed is that there always exists a matrix $A$ with the claimed properties that is similar to such a matrix $B$. However the question was whether every matrix with these properties is similar to such a matrix $B$. So let me also address this.
If $A^k=I$, then $\det(A^k)=\det(A)^k=\det(I)=1$. Now let $\lambda_1,\lambda_2$ denote the eigenvalues of $A$. Then, $\det(A)^n=(\lambda_1\lambda_2)^n=1$. Let us distinguish by the parity of $n$. If $n$ is odd, it follows that $\lambda_1\lambda_2=1$. If $n$ is even, it follows that $\lambda_1\lambda_2\in\{-1,1\}$.
Now if $A$ is similar to such a $B$, it needs to have the same eigenvalues and we know already how a $B$ similar to $A$ would have to look like. Since the determinant of $B$ is $$\det(B)=\cos(t)^2+\sin(t)^2=(\cos(t)+i\sin(t))(\cos(t)-i\sin(t))=1,\; t=\frac{2\pi}{n},$$ as soon as we choose such an $A$ with $\lambda_1\lambda_2=-1$, it cannot be similar to such a matrix $B$. (It is simple to write down examples, if you can't let me know and I will do so.)
Let us now assume that $\lambda_1\lambda_2=1$. By using polar coordinates on $\mathbb C$, $\lambda_1=\cos(\theta)+i\sin(\theta)$ and $\lambda_2=\cos(\theta)-i\sin(\theta)$ for some $\theta\in (0,2\pi]$. Now there is a result in linear algebra telling you that you find a base change such that $A$ becomes $$\begin{pmatrix}\cos(\theta)&\sin(\theta)\\ -\sin(\theta)& \cos(\theta)\end{pmatrix},$$ you can find it for example on wikipedia. This proves that if $\det(A)=1$, then it is indeed similar to such a matrix $B$. My original answer now even tells you what $\theta$ you'll get.
A: Hint If $A^n = I$, every eigenvalue $\lambda$ of $A$ must satisfy $\lambda^n = 1$, so that we can write $\lambda = e^{2 k \pi i / n}$ for some integer $k$. If both eigenvalues of $A$ are $\pm 1$, then $A^2 = I$, which, since $n \geq 3$, is excluded by the condition that $A^j \neq I$ for $0 < j < n$. So, $A$ has a nonreal eigenvalue, and since $A$ is real the other eigenvalue must therefore be its conjugate. Thus its eigenvalues are $e^{\pm 2 k \pi i / n} = \cos \frac{2 k \pi}{n} \pm i \sin \frac{2 k \pi}{n}$ for some integer $k$ not a multiple of $n / 2$. (In fact, the condition that $A^j \neq I$ for $0 < j < n$ implies that $k$ and $n$ are coprime.)
On the other hand, the eigenvalues of $B$ are $\cos t \pm i \sin t$.
