It seems that the following result should hold, but I can't find it explicitly anywhere. If $A=A^*$ is a properly supported pseudodifferential operator, does this imply that $\sigma_A(x,\xi)=\sigma_{A^*}(x,\xi)$? More specifically, is the symbol of $A$ real?
I know that the asymptotic expansion of $\sigma_{A^*}$ is given from the asymptotic expansion of $\sigma_A$ as $$ \sum_{\alpha}\partial^{\alpha}D^{\alpha}\overline{\sigma_{A}(x,\xi)}\cdot \frac{1}{\alpha} $$
This is true. As noted in the comments, it is (of course) true that if two operators are equal, then their principal symbols are equal. Otherwise, it would be ill-defined. Then, you clarified that you wanted to know if the principal symbol is necessarily real, and this is also true. The principal symbol of the adjoint is the complex conjugate of the principal symbol, i.e. $\sigma_{A^*}=\overline{\sigma_A}.$ Since $A^*=A,$ it follows that $\sigma_A=\sigma_{A^*},$ i.e. $\sigma_A=\overline{\sigma_A}.$ Thus, $$\text{Im }\sigma_A=\frac{1}{2i}\left(\sigma_A-\overline{\sigma_A}\right)=0,$$ implying that $\sigma_A$ is real. The full symbol need not be real, as the principal symbol is only defined up to a symbol of a lower order (i.e. if $A\in OPS^m,$ then $\sigma_A\in S^m/S^{m-1}$).