Two questions about commutative theories Let $\mathcal{T}$ be a commutative algebraic theory (for example sets, abelian groups, commutative monoids, but not groups etc.). References include the nlab and Borceux' Handbook of Categorical Algebra 2, section 3.10. Then $\mathsf{Mod}(\mathcal{T})$ is a monoidal category with internal homs.
Question 1. (Answered: Yes) Can we find a property of concrete categories which holds for $\mathsf{Mod}(\mathcal{T})$ if and only if $\mathcal{T}$ is commutative? In other words, does commutativity of an algebraic category not depend on the presentation? 
Question 2. (Answered: No) Let $\mathcal{T}$ be a commutative algebraic theory and $C=\mathsf{Mod}(\mathcal{T})$. Assume that $X \in C$ is a Co-$C$-algebra, i.e. we have a factorization of $\hom(X,-) : C \to \mathsf{Set}$ over $C$. Does this have to coincide with the usual factorization? This is well-known to be true in the examples I have mentioned above, for example for every abelian group $A$ there is only one natural abelian group structure on the hom-sets $\mathrm{hom}(A,B)$. 
This should be all well-known, but I don't know a reference.
 A: As for question 1: commutativity doesn't depend on the presentation of $T$. If $M = Mod(T)$ and $U: M \to Set$ is the forgetful functor, then commutativity can be formulated as saying that the monad $Ran_U U = U \circ Ran_U 1_M$ is commutative (or monoidal) in the sense of the nLab article here. Perhaps the most interesting aspect of this is that commutativity is a property, not an extra structure on a monad (where the structure of a strength constraint on an endofunctor on $Set$ is canonically given because every such endofunctor is canonically $Set$-enriched). These observations also lift to the enriched setting, provided of course that the functors involved are given as enriched functors (with respect to a base of enrichment $V$). 
(Note: $Ran_U 1_M$, which invariably exists, is just the left adjoint $F$ of $U$ if $U$ has a left adjoint. Some related discussion on the codensity monad of a general functor $U: M \to Set$ can be found in this post by Tom Leinster.) 
Edit: I had responded to question 2 earlier, but I am now editing that response out as it is superseded by Martin's second comment below, which makes the situation quite clear. Apologies for the noise. 
