In response to the edit: the action of the symmetric group $S_2$ on $E(2)$ induces an action on the corresponding summand of the free operad indexed by the tree you indicated. It might be helpful to think of elements of the free operad as formal operations $\theta$ generated from the $\mathbb{S}$-module; if you start with a ternary operation $\theta(x_1, x_2, x_3)$ in the summand corresponding to your tree, then you can define a new operation $\psi(x_1, x_2, x_3) = \theta(x_3, x_2, x_1)$, but this operation lies in the same summand [in the intended Loday-Vallete representation].

Yes, this edit looks right. Of course the usual tree notation may be easier to follow; the reason I used this set-theoretic notation is that I couldn't be bothered figuring out how to create and embed the graphics.

@D.S. Agree about the raft of suspicious answers, but this one could be legit, so I'm reposting in case it actually is useful.

I like the general thrust of the question, since it's asking for intuitions that often do not see the formal light of day. But I feel some inner resistance towards the formulation. The use of "at most" and "at least" looks backwards to me, because to me "at least" would refer to a minimal set of conditions, not a more stringent set. "It is generally agreed" is not something you can assert if you're getting your information "purely from a friend". As a category theorist, I'm not sure I agree that there's always a "right" notion of subobject for a given category. Depends on what you want to do.

Make sense? We have $\eta_Y g \eta_Y = \eta_Y$ using the fact that $g$ is a left inverse of $\eta_Y$. So $\eta_Y g \eta_Y = 1_Y \circ \eta_Y$. Now conclude $\eta_Y g = 1_Y$ since $\eta_Y$ is an epimorphism.

Oh, I see what you're trying to ask (I was thrown off by your notation). Since you've shown $\eta_Y$ has a left inverse, it will suffice to show it is an epimorphism. So suppose $f, g$ are distinct morphisms of the form $Y \to Y'$. Since the diagonal arrow in your diagram is injective, the two morphisms $f \eta_Y, g\eta_Y$ of the form $LRY \to Y'$ are also distinct. And this completes the proof.

When you say $R$ is fully faithful, that means the arrow you've denoted by $R$ is injective (that's the faithful part) and surjective (that's the full part). So this arrow $R$ is bijective: is invertible. So $\eta_Y$ is a composition of two isomorphisms.

It's only a difference in convention. See for instance ncatlab.org/nlab/show/dualizable+object#in_a_monoidal_catego‌​ry, especially Remark 2.2. The mathematics is the same; you modulate between these two definitions of dual by reversing the order of tensor factors. It's hard to enforce universal agreement with these sorts of conventions, just like the fact that some people compose functions in diagrammatic order and others in the Leibnizian order.

All this is also known to category theorists under the catchphrase "doctrinal adjunction". See for example the second corollary here. It also generalizes well from monoidal categories to bicategories.

No, you just have to keep in mind how you construct transposes in terms of the unit and counit. $^t(u(X^\vee))$ is by definition the composition $$G(X) \overset{\eta \otimes 1}{\to} F(X) \otimes F(X^\vee) \otimes G(X) \overset{1 \otimes u(X^\vee) \otimes 1}{\to} F(X) \otimes G(X^\vee) \otimes G(X) \overset{1 \otimes \varepsilon} \to F(X)$$ where $\eta: I \to F(X) \otimes F(X^\vee)$ is the "unit" for a monoidal dual pair, and $\varepsilon: G(X^\vee) \otimes G(X) \to I$ is the counit for another. This looks much simpler if you use string diagrams.