In cartesian coordinates, a rotation around the z axis is represented as:
$$\left( \begin{array}{ccc} \cos\phi & \sin\phi & 0\\ -\sin\phi & \cos\phi & 0\\ 0&0&1 \end{array} \right)$$ What is the analogue in spherical coordinates?
EDIT My change of coordinates is:
$x=r \sin\theta\cos\phi$
$y=r \sin\theta\sin\phi$
$z=r \cos\theta$
We wish to rotate a vector $\vec{r}$ to $R\vec{r}$ for a suitable rotation matrix $R$. I should make an important conceptual correction: the correct representation of $\vec{r}$ is $r\hat{r}$, not $r\hat{r}+\theta\hat{\theta}+\phi\hat{\phi}$ or whatever you were thinking of. In particular$$\begin{align}\hat{r}&=\sin\theta\cos\phi\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k},\\\hat{\theta}&=\cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k},\\\hat{\phi}&=-\sin\phi\hat{i}+\cos\phi\hat{j}\end{align}$$is our orthonormal basis. We can write these equations as $\hat{S}_A=T_{Aa}\hat{C}_a$ (I sum over repeated indices), with $\hat{C}_a$ ($\hat{S}_A$) the Cartesian (spherical) basis vectors, e.g. $T_{ri}=\sin\theta\cos\phi$. Now you just want to compute $R_{AB}=T_{Aa}R_{ab}(T^{-1})_{bB}$, or in matrix terms $R^\text{spherical}=TR^\text{Cartesian}T^{-1}$; I'll leave that to you.