The [nlab page](http://ncatlab.org/nlab/show/operad#definition_as_monoid_23) says 



> 
> A (Set-based) operad is a monoid in the monoidal category
>  $(Psh(ℙ),∘,I)$, where $ℙ$ is the category of $\sqcup\_{n\ge 0} S\_n$.
> 
> 
> 


The monoidal structure is given by the so called complicated **substitution product** $∘$, which is roughly defined as follows. 


The cardinal sum is a functor $ℙ\times ℙ\to ℙ$, it extends (co-continuously) to a **tensor product** on $Psh(ℙ)$ by the **Day Convolution**. Then one argues that 
$$\underline{\hom}(Psh(ℙ), Psh(ℙ))=Psh(ℙ)$$
where $\underline{\hom}$ is taking respect to the above tensor product. Then we transfer the obvious composition on $\underline{\hom}(Psh(ℙ), Psh(ℙ))$ to $Psh(ℙ)$, obtaining the substitution product. 


My question is that whether the substitution product itself arises from some functor $ℙ\times ℙ\to ℙ$ via the Day convolution? 


It seems we can define a *substitution product* on $ℙ$ by the genuine substitution operations.


Added. As pointed below, the substitution product is not bi-co-continuous. Now I should ask whether the substitution product restricts to representables, if yes, then how it extends to the $Psh(ℙ)$. 


The aim of my question is to understand the substitution product on the smaller space $ℙ$ as much as possible.



A question on the definition of operad
The answer is no, because any Day convolution product $F \ast G$ on presheaves $F, G: \mathbb{P}^{op} \to Set$ is cocontinuous in each of the separate arguments $F, G$, and yet the substitution product $F \circ G$ is not separately cocontinuous (only $- \circ G$ is cocontinuous, not $F \circ -$). 




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In case it is of some use to the OP: it is nevertheless interesting to consider what the restriction of the substitution product to representable presheaves yields, i.e., to calculate 


$$\mathbb{P} \times \mathbb{P} \stackrel{y \times y}{\to} \mathbf{Psh}(\mathbb{P}) \times \mathbf{Psh}(\mathbb{P}) \stackrel{\circ}{\to} \mathbf{Psh}(\mathbb{P})$$ 


where $y$ is the Yoneda embedding. The quick answer is nice and elegant; it's just 


$$\mathbb{P} \times \mathbb{P} \stackrel{\odot}{\to} \mathbb{P} \stackrel{y}{\to} \mathbf{Psh}(\mathbb{P})$$ 


where $\odot$ is the (symmetric) monoidal product on $\mathbb{P}$ which comes about by restricting the usual cartesian product on the category of finite sets and functions to the category of finite sets and bijections. This can be calculated in a variety of ways; one pleasant way of doing it is to identify the monoidal category $(\mathbf{Psh}(\mathbb{P}), \circ, I)$ with the monoidal category of analytic endofunctors on $\mathbf{Set}$ whose monoidal product is composition of functors; for this, see the appendix to Joyal's article 


* A. Joyal, Foncteurs analytiques et espèces de structures, Lecture Notes Math. 1234, Springer 1986, 126–159.


Here the analytic functor corresponding to a representable $\mathbb{P}(-, p)$ is just $x \mapsto x^p$, the $p^{th}$ power functor. Since the composite of two such power functors is given by $(x^p)^q \cong x^{p \odot q}$, we get the result. 


This actually sheds another light on why substitution *cannot* come from a Day convolution: this calculation shows that if it did, then it could only be the Day convolution induced from $\odot: \mathbb{P} \times \mathbb{P} \to \mathbb{P}$. As this is a symmetric monoidal product, the induced Day convolution would also be symmetric monoidal. But the substitution product is highly non-symmetric: $F \circ G \cong G \circ F$ is of course generally false, just as composition of (analytic) endofunctors is non-symmetric! 


