Conjugation by rotations acts on w by e^{2iβ}

Idea. Rotations commute with the conjugate-linear part only up to a phase. The complex-linear part $x\mapsto zx$ commutes with all rotations, so $z$ is unchanged.

Computation. For $x\in\mathbb{C}$, $R_{-\beta}x=e^{-i\beta}x$, so
$$AR_{-\beta}x=z\,e^{-i\beta}x+w\,\overline{e^{-i\beta}x}=z\,e^{-i\beta}x+w\,e^{i\beta}\bar x.$$
Applying $R_\beta$ (multiply by $e^{i\beta}$):
$$R_\beta AR_{-\beta}x=zx+w\,e^{2i\beta}\bar x.$$
By uniqueness of the decomposition $Ax=zx+w\bar x$ (the map $A\mapsto(z,w)$ is bijective), the coordinates of $R_\beta AR_{-\beta}$ are $(z,e^{2i\beta}w)$.

Check (numerical). For random $A$ and random $\beta$, the coordinates of $R_\beta AR_\beta^T$ agree with $(z,e^{2i\beta}w)$ to $10^{-15}$.

Consequences.
- The quantities $z$, $|w|$ (hence $\operatorname{tr}A=2\operatorname{Re}z$, $\det A=|z|^2-|w|^2$, $\|A\|_F$) are invariant under conjugation by $SO(2)$.
- $A$ commutes with every rotation iff $w=0$, i.e. iff $A$ is a "complex scalar" $x\mapsto zx$, equivalently $A=\begin{pmatrix}a&-c\\c&a\end{pmatrix}$.
- Conjugating by the reflection $x\mapsto\bar x$ gives $\overline{A\bar x}=\bar zx+\bar w\bar x$: $(z,w)\mapsto(\bar z,\bar w)$.

Remark. The factor $e^{2i\beta}$ (and not $e^{i\beta}$) shows that the rotation $R_\pi=-I$ acts trivially, as it must, since $-I$ is central; and that $R_\beta$ and $R_{\beta+\pi}$ give the same conjugation.

Lesson. To understand a group action on a space of matrices, find coordinates in which it becomes diagonal (here: multiplication by phases).