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Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings. Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$.

Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course there are non-isomorphic $G$-sets whose underlying sets are isomorphic.