Free operads and trees I am currently working in my PhD thesis, and it became necessary to understand some facts about free symmetric operads. Hence I started to study this subject by myself, following Kapranov & Ginzburg's paper (Koszul Duality for Operads) and the Bresse's paper (Koszul Duality of Operads and Homology of Partition Posets). Naturally, a few doubts has arisen. Since I had a couple of questions on this topic unanswered in math stackexchange, I will post my new questions here - they are not so trivial to me, but I think that they are probably routine for researchers.
My first question concerns the intuition about this topic. Usually in algebra, the operations in free structures are described via concatenation, e.g., on free groups, on free algebras, etc. Keeping this in mind, I thought about the operad product (on free symmetric operads) as a concatenation via tensor products, where the trees play a similar role in operads as the "reduction rules" in free groups, in order to guarantee that the axioms of the operations are satisfied. This analogy makes any sense?
Now, my second (and more technical) question is about the construction of a free symmetric operad, as done here (pages 69 to 71). What exactly are the elements of the $\Sigma_n$-module $F(M)(n)$? Also, how to explicitly  describe (i.e., in elements) the operad product on $F(M)$?
 A: This is an answer to your second question.
An element of consists of the equivalence class of a pair $(\tau,x)$. Here $\tau$ is a tree with $n$ leaves and $x$ is an element of $\tau(M)$. We'll unpack this in a second. The equivalence relation comes from isomorphisms of trees. There is an easy kind of tree and a hard kind of tree in terms of isomorphisms so let's start with the easy kind of tree first.
The easy kind of tree is the kind with only the identity automorphism. An easy (and common) criterion that is sufficient (though not necessary) for a tree to have only the identity automorphism is that every vertex has at least one incoming edge. For example, these will be the only trees that contribute to the free operad if the $\Sigma_*$ module $M$ has $M_0=0$.
In this easy case case, we can choose any representative tree (say that we choose one $\tau_0$ where the vertices are named $v_1,\ldots, v_r$ and the edges are named $e_1\,\ldots, e_{r+n}$). Then we can represent the equivalence class by an element 
$$
(\tau_0, x_0)
$$
where $x_0$ is an element of 
$$M(I_{v_1})\otimes\cdots \otimes M(I_{v_r})$$
(recall that $I_i$ is the number of incoming edges of the vertex $v_i$) and each such pair is alone in its equivalence class.
On the other hand, if a tree has automorphisms, then we must further mod out by these automorphisms. We can still pick a representative $(\tau_0,x_0)$ as above, but the pair is not necessarily alone in its equivalence class. The automorphism induces an automorphism $\phi$ on the vertices $V(\tau_0)$ and for each vertex $v_i$, an isomorphism $\phi_i:I_{v_i}\to I_{\phi(v_i)}$. Then
$$m_1\otimes\cdots \otimes m_r \in M(I_{v_1})\otimes\cdots \otimes M(I_{v_r})
\sim
\phi_1^{-1}(m_{\phi^{-1}(v_1)})\otimes\cdots\otimes \phi_r^{-1}(m_{\phi^{-1}(v_r)})
$$
The easiest example is the $0$-tree

where the relation is $$(\tau_0,m_1\otimes m_2\otimes m_3)\sim (\tau_0, m_2\otimes m_1\otimes \sigma(m_3)$$ in the ungraded case and
$$
(\tau_0,m_1\otimes m_2\otimes m_3)\sim (-1)^{|m_1||m_2|}(\tau_0, m_2\otimes m_1\otimes \sigma(m_3)
$$
in the graded case. In both cases $\sigma$ is the nontrivial element of $\Sigma_2$, which acts on $m_3\in M(2)$.
This should fit with an intuition about operads that comes from algebra, because if you have two elements $a=m_1$ and $b=m_2$ in an algebra (of whatever sort) with a binary product $\mu m_3$, then you might expect $\mu_3(a,b)$ to agree with $\mu_3^{op}(b,a)$, and since you expect this to be universally true regardless of any properties of $a$, $b$, and $\mu$, it should be true in the free operad.
