Given a small set $S$, we can define the overcategory $\mathbf{Set}/S$ to be the category whose objects are pairs $(A:\mathbf{Set},a:A\to S)$ and whose morphisms $(A,a)\to(B,b)$ are functions $f:A\to B$ with $a = bf$. We can also define the functor category $\mathbf{Set}^S$ by treating $S$ as a discrete category. From these definitions one can prove that we have an equivalence $\mathbf{Set}/S\simeq\mathbf{Set}^S$.
I'm wondering if this result can be categorified. Let $\mathbf{Cat}$ be the $2$-category* of small $1$-categories, and let $\mathcal C$ be a small $1$-category. We may define $\mathbf{Cat}/\mathcal C$ and $\mathbf{Cat}^{\mathcal C^\mathrm{op}}$ analogously to the above. The use of $\mathcal C^\mathrm{op}$ is needed to make sure these two constructions have the same variance in $\mathcal C$.
Is $\mathbf{Cat}/\mathcal C\simeq\mathbf{Cat}^{\mathcal C^\mathrm{op}}$? If not, is there a simple counterexample?
*Assume everything is "weak".
I know that this is related to the Grothendieck construction, which shows that $\mathbf{Cat}^{\mathcal C^\mathrm{op}}$ is equivalent to the $2$-category of fibrations over $\mathcal C$. Since this is a subcategory of $\mathbf{Cat}/\mathcal C$, this suggests that $\mathbf{Cat}^{\mathcal C^\mathrm{op}}$ isn't also equivalent to $\mathbf{Cat}/\mathcal C$. But it doesn't rule out that there might be an equivalence via some other construction. So I'd prefer that we could prove that $\mathbf{Cat}/\mathcal C$ and $\mathbf{Cat}^{\mathcal C^\mathrm{op}}$ aren't equivalent by producing some simple $\mathcal C$ for which they have different properties. Letting $\mathcal C$ be a set doesn't work, because $\mathbf{Cat}/\mathcal C$ and $\mathbf{Cat}^{\mathcal C^\mathrm{op}}$ really are equivalent in this case.
Take $\cal C$ equal to the ordinal $[2]$. The slice is not cartesian closed, while the functor category is (because $\bf Cat$ is). See the notion of Conduché functor on the nLab (where the ordinal [2] is called "3", alas!)