rotate a coordinate frame's position and orientation together say we have a coordinate frame defined by T_po, w.r.t. the world frame.
How can I rotate this coordinate frame using a rotation matrix such both position and orientation rotate together. So far I tried multiplying a rotation matrix, but this rotation matrix only rotates the frame locally with respect to it's own frame.

 A: If I understand your statements correctly, you have the homogeneous transformation $T^O_A$ relating coordinate frame $A$ to the global reference frame, and you wish to compute $T^O_B$, generated by rotating the frame $A$ by an angle $\theta$ about the origin.

Let us represent $T^O_A$ as:
$$
T^{O}_{A} =  \begin{bmatrix}R^{O}_{A}&
 \begin{bmatrix}x_A\\y_A\end{bmatrix} 
 \\\mathbf{0}&1\end{bmatrix} 
$$
Realize that $R^{O}_{B} = R^{O}_{A}Rot(z,\theta)$, where $Rot(z,\theta) = \begin{bmatrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{bmatrix}$ is the rotation matrix about the z axis. To describe the origin of the frame $B$, we need to rotate the vector $\mathbf{r} = \begin{bmatrix}x_A\\y_A\end{bmatrix}$ by $Rot(z,\theta)$:
$$
\begin{bmatrix}x_B\\y_B\end{bmatrix} = Rot(z,\theta)\begin{bmatrix}x_A\\y_A\end{bmatrix} 
$$
Now, you can formulate $T^O_B$ as:
$$
T^{O}_{B} =
\begin{bmatrix}R^{O}_{B}&
 \begin{bmatrix}x_B\\y_B\end{bmatrix} 
 \\\mathbf{0}&1
\end{bmatrix} 
=
\begin{bmatrix}R^{O}_{A}Rot(z,\theta)&
 Rot(z,\theta)\begin{bmatrix}x_A\\y_A\end{bmatrix} 
 \\\mathbf{0}&1
\end{bmatrix} 
$$
A: The original frame has its origin at $d_A$ and has its axes as columns of the rotation matrix $R_A$, then the global (world) coordinates are given by
$ p_W = d_A + R_A p_A $
Now you want to rotate $p_W$ about the origin through an angle $\theta$, which generates the rotation matrix $R$
$R = \begin{bmatrix} \cos( \theta) && - \sin (\theta) \\ \sin( \theta) && \cos(\theta) \end{bmatrix} $
and the new coordinate vector after rotation is
$p'_W = R p_W = R d_A + R R_A p_A  = d'_A + R'_A p_A $
with obvious correspondence between the variables.
