Let $f\colon\mathbb{R}^n\rightarrow\mathbb{C}$ be a bounded function with bounded first derivative. Then the multiplication operator $$H^1(\mathbb{R}^n)\rightarrow H^1(\mathbb{R}^n),$$ $$s\mapsto A_f x:=fx$$ is bounded, where $H^1(\mathbb{R}^n)$ is the Sobolev space $W^{1,2}(\mathbb{R}^n)$.
In two previous questions (see here and here), it was asked whether the adjoint $A_f^*$ is also given by function multiplication. This answer by Iosif Pinelis shows that this is the case if and only if $f$ is a constant function.
The following are some further questions about the nature of $A_f^*$.
Q1: Does $A_f^*$ preserve the space $C_c^\infty(\mathbb{R}^n)$?
Q2: Is $A_f^*$ a pseudodifferential operator?
Q3: Would the answer to either of the above change if $\mathbb{R}^n$ is replaced by a compact manifold?
Let $\overline{f}$ denote the complex conjugate of $f$, and let $A_{\overline{f}}$ be the associated multiplication operator. For $g,h\in C_c^\infty(\mathbb{R}^n)$, we have: \begin{align} \langle g, A_fh\rangle_{H^1}&=\langle g,(1+\Delta)A_fh\rangle_{L^2}\\ & = \langle A_{\overline{f}}(1+\Delta)g,h\rangle_{L^2}\\ & = \langle (1+\Delta)^{-1}A_{\overline{f}}(1+\Delta)g,h\rangle_{H^1}. \end{align}
Thus $A_f^*=(1+\Delta)^{-1}A_{\overline{f}}(1+\Delta)$, a pseudodifferential operator. So the answers to the first two questions are both "yes". Since the proof works equally well on a manifold, the answer to the third question is "no".