S-modules and Schur functors I am reading the book "Algebraic Operads" by Loday and Vallette.
(I will refer to their version 0.999 here : http://math.unice.fr/~brunov/Operads.pdf)
In Chapter 5, they define an $\mathbb{S}$-module as a collection
$$(M(0),M(1),...)$$
where $M(n)$ is an $k[S_n]$-module.
A morphism of $\mathbb{S}$-module is a collection of $k[S_n]$-homomorphism $M(n) \to N(n)$
Give an $\mathbb{S}$-module $M$, the associated Schur functor is the functor $\mathsf{Vect}_k \to \mathsf{Vect}_k$ defined as
$$V \mapsto \bigoplus_n M(n)\otimes _{k[S_n]}V^{\otimes n}$$
Reading that chapter, I have the impression that they want to make the following claim (but have never bothered to make it explicit):
$\textbf{Claim}$ The functor described above:
$$ \mathbb{S}\text{-modules} \to EndoFunc(\mathsf{Vect}_k)$$
is fully-faithful.
Is this claim true? If yes, is there reference in which it is written down? Or could someone sketch a proof? Also, is this true as well if we replace $\mathsf{Vect}_k$ by any symmetric monoidal category in which the monoidal product commute with coproducts? (see the remark on page 105 where it is claimed that the notion of operads can be defined on any such monoidal category)
In the book, there is the Lemma 5.1.3, which seems to suggest we can recover the $S_n$-module $M(n)$ from the associated Schur functor by applying $V=ke_1\oplus ...\oplus ke_n$ and "take the multilinear part". But I don't know how the multilinear part can be obtained from merely the data of an endofunctor. Any idea?
 A: I do not think it is full. For example let $M = \mathtt{Com}$ the commutative operad defined by $M(n) = k$ (with the trivial $\mathbb{S}_n$ action) for all $n > 0$ and $M(0) = 0$. The associated Schur functor is the free symmetric (nonunital) algebra functor $S = \widetilde{\mathtt{Com}}$.
There is a natural transformation $\alpha : S \to S$, $\alpha_V : S(V) \to S(V)$ given by $\alpha_V(x) = x^2$. This is a natural transformation, because if $f : V \to W$, the way $S(f)$ is defined implies that it commutes with squaring. As pointed out to me by Mario Gonçalves Lamas, this counter-example requires characteristic $2$ to make sure $\alpha_V$ is linear.
But there is no endomorphism $\lambda : M \to M$ such that $\tilde\lambda = \alpha$. Indeed such an endomorphism is of the form $\lambda_n : M(n) \to M(n), \lambda_n(x) = \lambda_n' x$ for some constants $\lambda_n'$. So if eg. $V = k \langle x \rangle$ has dimension one, $S(V) = k[x]$ and the morphism $k[x] \to k[x]$ induced by $\lambda$ is given by $x^k \mapsto \lambda_k' x^k$. This cannot be $\alpha$.

It is, however, faithful. Suppose $f, g : M \to N$ are two $\mathbb{S}$-module morphisms such that
$$\tilde{f} = \tilde{g} : \widetilde{M} \to \widetilde{N}.$$
Then for all $n$, consider $V = k \langle x_1, \dots, x_n \rangle$ a space of dimension $n$. Then in the spirit of the Lemma 5.1.3 (actually Lemma 5.1.1 in the published version of the book), the $n$-multilinear part of $\widetilde{M}(V)$ is isomorphic to $M(n)$, that of $\widetilde{N}(V)$ to $N(n)$, and the restriction of $\tilde{f}$ to that part is given, through this isomorphism, by $f_n : M(n) \to N(n)$. But since $\tilde{f} = \tilde{g}$, it follows that $f_n = g_n$ for all $n$, so $f = g$.
A: For the sake of completness, I would like to point out that Najib's counter-example requires a bit more than a condition on the field's characteristic. Linearity of $\alpha_V$ implies the relations $(x+y)^2=x^2+y^2$ and $(\lambda x)^2=\lambda x^2$, and those can only be true if we work with the field $\mathbb{F}_2$.
This example can be adapted to any finite field, but when the field $k$ is infinite it can be proved that the functor
$$S\colon\mathbb{S}\text{-}mod \longrightarrow Endo(\mathrm{Vect}_k)$$
is fully faithful.
This result is mentioned by Benoit Fresse in Modules over Operads and Functors, at the beginning of section 2.3. 
The proof is based on the existence of a right adjoint $\Gamma\colon Endo(\mathrm{Vect}_k)\rightarrow\mathbb{S}\text{-}mod$. If $M,N$ are $\mathbb{S}$-modules, the morphism $Hom_{\mathbb{S}\text{-}mod}(M,N)\overset{S}{\longrightarrow} Nat(SM,SN)$ can be described as the composite
$$ Hom_{\mathbb{S}\text{-}mod}(M,N) \overset{\eta(N)_*}{\longrightarrow} Hom_{\mathbb{S}\text{-}mod}(M,\Gamma SN)\simeq Nat(SM,SN)$$
where $\eta(N)\colon N\rightarrow\Gamma SN$ is the unit of the adjunction.
The proof that $\eta(N)$ is indeed an isomorphism when $k$ is an infinite field requires subtle arguments from representation theory, Benoit Fresse gives another book as a reference : Symmetric Functions and Hall Polynomials by I. G. Macdonald.
