If $D$ is a stable monoidal $\infty$-category then the theory of Smith (co)ideals tells you that if you have a monoidal functor $F: \mathbb{Z}_{\leq 0} \to D$ then the fiber of the map $F(-1\to 0)$ is a monoid and every monoid arises this way.

If you take opposites this should tell you that a well pointed endofunctor must be at least as complicated as a monoid.

If $D$ is a stable monoidal $\infty$-category then the theory of Smith ideals tells you that if you have a monoidal functor $F: \mathbb{Z}_{\leq 0} \to D$ then the cofiber of the map $F(-1\to 0)$ is a monoid and every monoid arises this way.

If you take opposites this should tell you that a well pointed endofunctor must be at least as complicated as a comonad.

If $\text{End}(C)$ is a stable $\infty$-category then the theory of Smith (co)ideals tells you that the cone on $\sigma$ is a comonad and every comonad on C arises this way. Thus presenting well pointed endofunctors must be at least as hard as presenting comonads on stable $\infty$-categories.

If $D$ is a stable monoidal $\infty$-category then the theory of Smith (co)ideals tells you that if you have a monoidal functor $F: \mathbb{Z}_{\leq 0} \to D$ then the fiber of the map $F(-1\to 0)$ is a monoid and every monoid arises this way.

If you take opposites this should tell you that a well pointed endofunctor must be at least as complicated as a monoid.