Generalized adjoint operation valid? Let $R_{\theta}$ be the rotation by an angle $\theta$.
Is it then true that for multi-indices $\alpha$ of fixed order $j$ and any smooth function $f$ we have
$$\sum_{\vert \alpha \vert=j}(R_{\theta}z)^{\alpha} \partial^{\alpha}f(x) = \sum_{\vert \alpha \vert=j}(z)^{\alpha} (R_{-\theta}\partial)^{\alpha}f(x)$$
It is true for $j=1$, which just follows since $R_{-\theta}$ is the adjoint map of $R_{\theta}.$
 A: If I am understanding your notation correctly (which admittedly, I may not be), I believe this is already false for $j = 2$. Let me write:
$$ R_{\theta} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \qquad f = f(x,y) \qquad z = \begin{bmatrix} u \\ v \end{bmatrix} $$
Then for $j = 2$, the LHS is:
\begin{align*}
\text{LHS} &=
    (u\cos\theta - v\sin\theta)^2 \frac{\partial^2 f}{\partial x^2} \\
    &\hspace{1cm} + (u\cos\theta - v\sin\theta) (u\sin\theta + v\cos\theta) \frac{\partial^2 f}{\partial x \partial y} \\
    &\hspace{1cm} + (u\sin\theta + v\cos\theta)^2 \frac{\partial^2 f}{\partial y^2}
\end{align*}
and the RHS is:
\begin{align*}
\text{RHS} &=
    u^2 \left(
        \cos^2\theta \frac{\partial^2 f}{\partial x^2}
        + 2\sin\theta\cos\theta \frac{\partial^2 f}{\partial x \partial y}
        + \sin^2\theta \frac{\partial^2 f}{\partial y^2}
    \right) \\
    &\hspace{1cm} + uv \left(
        -\sin\theta \cos\theta \frac{\partial^2 f}{\partial x^2}
        + (\cos^2\theta - \sin^2\theta) \frac{\partial^2 f}{\partial x \partial y}
        +\sin\theta \cos\theta \frac{\partial^2 f}{\partial y^2}
    \right) \\
    &\hspace{1cm} + v^2 \left(
        \sin^2\theta \frac{\partial^2 f}{\partial x^2}
        - 2\sin\theta \cos\theta \frac{\partial^2 f}{\partial x \partial y}
        + \cos^2\theta \frac{\partial^2 f}{\partial y^2}
    \right)
\end{align*}
If you fully expand out the products, you will find that these two quantities differ by:
$$ \text{LHS} - \text{RHS} = \sin\theta\cos\theta \left(
    (v^2-u^2) \frac{\partial^2 f}{\partial x \partial y}
    - u v \left(\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2}\right)
\right) $$
This is certainly not identically zero, so $\text{LHS} \ne \text{RHS}$ in general.
