If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right) A + \det A$.
Since $-A^3 + \det A \cdot I = 0$, then $A = \frac{\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right)}{\text{tr} A}I$. With this in mind I cannot obtain even rotation matrix to say nothing about something else. Any hints?

Since $A^3-I=0$ you get that the possible eigenvalues are $1, e^{2\pi i/3}, e^{4\pi i/3}$. 
Moreover, since $A$ is $3 \times 3$ it has 1 or 3 real eigenvalues.
Case 1: All eigenvalues are real. Then, the minimal polynomial of $A$ is $(x-1)^k$ and divides $x^3-1$, therefore it is $x-1$. This shows that $A$ is diagonalizable and hence
$$A=PIP^{-1}=I$$.
Case 2: $A$ has two complex eigenvalues.  Then, $1, e^{2\pi i/3}, e^{4\pi i/3}$ are all eigenvalues, with eigenvectors $u,v, \bar{v}$.
You have $A=PDP^{-1}$ where 
$$P=(u v \bar{V})$$ and $D=diag (1, e^{2\pi i/3}, e^{4\pi i/3})$
Now, write $$v=v_1+i v_2$$ Show that 
$$A= (u \;v_2 \;v_2) \begin{bmatrix}1 &0 & 0\\
0& \cos(2\pi /3 )& \sin(2\pi /3)\\
0& -\sin(2\pi /3 )& \cos(2\pi /3)
\end{bmatrix}(u \;v_2 \;v_2) ^{-1}$$
