Let $\mathcal{O}$ be an operad in the monoidal category $M$. Then $\mathcal{O}(1)$ together with the morphisms $$\mathcal{O}(1)\otimes \mathcal{O}(1)\to \mathcal{O}(1)$$ and the unit $\eta:1\to \mathcal{O}(1)$ is a monoid object. Moreover, a morphism $\varphi:\mathcal{O}\to \mathcal{O}'$ of operads induces a morphism $\mathcal{O}(1)\to \mathcal{O}(1)$ of monoid objects. Therefore, we get a forgetful functor $$\mathrm{Operads}(M)\to \mathrm{Monoids}(M).$$ If we work with coloured operads, we get a functor from coloured $M$-operads to $M$-enriched categories. Conversely, if $T$ is a monoid object, we can build an operad by $$\mathcal{O}_T(r) := T^{\otimes r}.$$ and the structure maps $$T^{\otimes r}\otimes \bigotimes_{i=1}^r T^{\otimes k_i}\to T^{\otimes (k_1+\dotsb+k_r)}$$ as follows: Let $\Delta:T\to T^{\otimes k}$ be the diagonal (existence is clear if $\otimes$ is the categorical product). Then $$T\otimes T^{\otimes k}\stackrel{\Delta\otimes \mathrm{id}}{\to}T^{\otimes k}\otimes T^{\otimes k} \cong (T^{\otimes 2})^{\otimes k} \to T^{\otimes k}.$$ In $\mathbf{Set}$, this just means $t(t_1,\dotsc,t_k)=(tt_1,\dotsc,tt_k)$. It should be clear that this construction gives us an operad. Now two problems/questions:

- Does the morphism $\Delta$ always exist in the general setting? It is obviously not the same as $$T\cong T\otimes 1^{\otimes (k-1)}\stackrel{\mathrm{id}\otimes \eta^{\otimes (k-1)}}{\to} T^{\otimes k}.$$
- Obviously $\mathcal{O}_T(1)\cong T$, but it seems to be not true that $\mathcal{O}_T$ is the free operad over the monoid object $T$. Is there another construction for the “free” operad over $T$?