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:


1. Does the morphism $\Delta$ always exist in the general setting? It is obviously not the same as
$$T\cong T\otimes 1^{\otimes (k-1)}\stackrel{\mathrm{id}\otimes \eta^{\otimes (k-1)}}{\to} T^{\otimes k}.$$
2. 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 $T$?



Free operad over a monoid object
Let me mention that this is related to [this earlier question of mine](https://mathoverflow.net/q/254714/36146) (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $\mathtt{Com} \rtimes T$ of the commutative operad $\mathtt{Com}$ by $T$.


1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $\mu : T \otimes T \to T$ and a comultiplication $\Delta : T \to T \otimes T$ such that $\mu$ is associative and unital, $\Delta$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $\Delta \circ \mu = \mu \circ (\Delta \otimes \Delta)$.
2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $\varnothing$ be the initial object of your category and suppose that $\varnothing \otimes X = \varnothing = X \otimes \varnothing$ for all $X$. Then (as Simon Henry mentions in the comments), the free operad on $T$ is simply given by $\mathtt{O}(1) = T$ and $\mathtt{O}(n) = \varnothing$ for $n \neq 1$.


