Does Covariance matrix of a rotated data have the information of angle of rotation? Does covariance of a data rotated by some angle to the orginal data,have the information of rotation.
i.e I read in this article(http://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/) (assuming S as an Identity matrix in the link) that covariance of the rotated data is nothing but 
\begin{equation*}  \Sigma = R \, R^{-1} \end{equation*}
 A: Let $\Sigma$ be your covariance matrix for a random vector $\mathbf{x}$.
Now let's consider an eigenvalue decomposition of $\Sigma$ such $Q\Lambda Q' = \Sigma$ where $\Lambda$ is a diagonal matrix and $Q$ is an orthogonal matrix (i.e. $QQ'=I$). The intuitive interpretation of $Q$ is that it is a rotation matrix (or an improper rotation matrix) and that $\Lambda$ acts a scaling matrix.
Let $\mathbf{z} = \sqrt{\Lambda^{-1}} Q' \mathbf{x}$. That is, we rotate $\mathbf{x}$ using $Q'$ and scale it by the square root of the inverse of $\Lambda$. Then:
$$\mathrm{Cov}\left(\mathbf{z} \right) = I$$
The different elements of the random vector $\mathbf{z}$ are orthogonal and have variance = $1$.
Alternatively, we could go the other direction. We could start with an orthogonal random vector $\mathbf{z}$ and obtain a random vector with covariance matrix $\Sigma$ by $\mathbf{x} = Q\sqrt{\Lambda} \mathbf{z}$. Observe that:
$$\begin{align*} \mathrm{Cov}\left( \mathbf{x} \right) &= \mathrm{Cov}\left( Q\sqrt{\Lambda} \mathbf{z} \right) \\
&= Q \sqrt{\Lambda} \mathrm{Cov}\left( \mathbf{z} \right) \sqrt{\Lambda}Q' \\
&= Q \sqrt{\Lambda} I \sqrt{\Lambda} Q' \\
&= Q \Lambda Q' \\
&= \Sigma
\end{align*}$$
You can do this as long as $\Sigma$ is positive definite, that is, the elements of your random vector $\mathbf{x}$ aren't linearly dependent. 
Recap:
You can go back and forth between a random vector $\mathbf{z}$ where $\mathrm{Cov}(\mathbf{z}) = I$ and a random vector $\mathbf{x}$ where $\mathrm{Cov}(\mathbf{x}) = \Sigma$ using $Q$ and $\Lambda$ obtained from an eigenvalue decomposition of $\Sigma$. $Q$ and $\Lambda$ are uniquely defined for a positive definite covariance matrix $\Sigma$. If you rotate and scale $x$ to obtain some $z$ then toss $\Sigma$ and $Q$, there's no magic way to go back.
