A generalization of holomorphic functions Let's fix a  matrix $A\in M_{2}(\mathbb{R})$.  Assume that the  following vector space of  smooth  functions is  closed under complex  multiplication:
$$\mathcal{S}_{A}=\{f:\mathbb{C}\to \mathbb{C}\mid Df.A=A.Df  \}$$ 
Here  $Df$  is  the  Jacobian of  $f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ (We  identify  $\mathbb{C}$  with $\mathbb{R}^{2}$).
Does  this  imply that  $A$ is  in the  form $A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$?
Note that For $A=\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}$  the  relation $Df.A=A.Df$  is equivalent to the Cauchy Riemann equations for $f=u+iv$ so we obtain the class of  holomorphic  functions.
 A: You're right. But the A is always satisfying Cauchy–Riemann equations.
Let $f = u(x,y)+v(x,y)i$,
By the definition of smoothness: the f must be differentiable which implies f must satisfy Cauchy–Riemann equations.
For simplification: denote $$\frac{\partial{f}}{\partial{x}} = f_x$$
By Cauchy–Riemann equations: $$u_x = v_y   \tag{1}$$ and $$u_y = - v_x \tag{2}$$
The Jacobian of f is $\begin{pmatrix} u_x&u_y\\v_x&v_y \end{pmatrix}$ Designate it as B.
Let A = $\begin{pmatrix} a&b\\c&d\end{pmatrix}$
Denote $U[i][j]$ is the entry on $i^{th}$ row $j^{th} $ column where U is a matrix;
By $A\times B =B\times A $
Then $(A\times B)[1][1] = au_x+bv_y=(B\times A)[1][1]  = au_x+cu_y $
By $(2)$,$ au_x+bu_y = au_x-bu_y$ forall $f$.
Thus $c = -b$.
Choose $(A\times B)[1][2] = (B\times A)[1][2] $ and using $(1)$, one can get d = a.
Choose  $(A\times B)[2][1] =  (B\times A)[2][1] $ one can show it has holed.
Choose  $(A\times B)[2][2] =  (B\times A)[2][2] $ one can show it has holed.
Thus A = $\begin{pmatrix} a&b\\-b&a\end{pmatrix}$
A: Put $f(z)=z$ Then $f$ belongs to $\mathcal{S}_{A}$. So $f^{2}$ must belong to $\mathcal{S}_{A}$,too.
This means that $A$ commutes with all rotations. Hence $A$ has the desired representation as it is pointed out in the comment by user 1952009.
