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It seems to me not, since this would seem to imply that for all functors $F$ and all objects $A$ in $C$ there exists a morphism $F(A) \to A$ (making all functors co-pointed?). However, intuitively it seems like the identity functor acts like a terminal object; a monad $M$ on $C$ is a monoid on $[C, C]$ where the "unit" is a natural transformation $η : I \to M$, while for a monoidal set $S$ in Set the unit is a function $e : 1 \to S$. So am I misunderstanding something, or are my intuitions leading me astray?

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The difference between those two examples is that in $\textbf{Set}$ the monoidal operation is the categorical product (so the identity object is the terminal object), whereas this is not true in the category of endofunctors on $\mathbf{C}$. (I believe the latter has a product if and only if $\mathbf{C}$ does, and then it is the pointwise product. It follows that the terminal object, if it exists, is the functor which sends all objects to $\mathbf{1}$ and all morphisms to the unique morphism $\mathbf{1} \to \mathbf{1}$. In particular, it's not the identity functor.)

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    @pelotom: yes. A monoidal structure on a category doesn't have to be the categorical product; it can be the categorical coproduct or something weirder (for example the tensor product on vector spaces).2010-10-08