The answer is yes. Stein operator is essentially no more than a differential equation $E$(written as an operator notation) which characterized a distribution $f$ as its unique solution. That is the reason why Stein operator method allows us to characterize distribution $f_Z$ as a unique solution to $E$. Consider the classic Stein operator $$E:f\mapsto f'(x)-xf(x)$$ and we can choose any $g:\mathbb{R}\rightarrow\mathbb{R}$ which is a bounded continuous function, then the Green function (simultaneously density function of a normal distribution)

$$f(x)=e^{\frac{x^{2}}{2}}{\displaystyle \int_{-\infty}^{x}(g(t)-\boldsymbol{E}_{f_Z}g(Z))e^{\frac{-t^{2}}{2}}dt}$$

is the unique bounded functional solution of the following ODE

$$f'(x)-xf(x)=g(x)-\boldsymbol{E}_{f_Z}g(Z)$$

Therefore the question to ask is what kind of $g$ you are looking for to characterize the distribution $f_Z$ using distribution properties of $g(Z),Z\sim f_Z$ to construct something on RHS and that is also why Stein operator is not unique, people usually find it hard to decide $g(x)$ in various applications due to the available measurement methods.

For example, if we know some information about the fourth moment of $f_Z$, then I may choose a Stein operator $E_1:f\mapsto f^{(4)}$ and then figure out what kind of $g$ will serve this purpose on the RHS of the equation $Ef= G(\boldsymbol{E}_{f_Z}{Z^4})$. On the other hand if we know what kind of Stein equation we must have, we can equivalently set up the boundary condition of a certain differential equation on the fourth derivative of the characteristic function of $f$ and then seek a solution to theis equation.

Reference

Chen, Louis HY, Larry Goldstein, and Qi-Man Shao. Normal approximation by Stein’s method. Springer Science & Business Media, 2010.