For a category $\mathcal{C}$ and $*$ a terminal object of it, let $\mathcal{C}_*$ denote the category under this terminal object.
Suppose that the opposite of $\mathcal{C}$ is $\mathcal{C}^{\text{op}}\simeq \mathcal{D}$. Is it true that $\mathcal{C}^{\text{op}}_*\simeq \mathcal{D_*}$?
Not quite-you have to switch to $D$ over $*$. Given any $X$ in a category $C$, $(X/C)^{op}$ has object the maps $X\to c$ in $C$ and morphisms $(X\to c)\to(X\to c')$ maps $c'\to c$ in $C$ making the triangle commute in $C$. Alternatively, write the objects as maps $c\to X$ in $C^{op}$ and similarly for the morphisms, and you have an isomorphism $(X/C)^{op}\cong C^{op}/X$.