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As for question 2

Edit: I haven't nailed down all the details, but if $G$ is the lift of $\hom(X, -): M \to Set$ provided by a co-$T$-algebra structure on $X$ internal to $M$, then provided that the canonical natural map

$$\hom_M(A, B) \had responded to \hom_M(GA, GB)$$

is an algebra map (as I think it will bequestion 2 earlier, but these are the details I haven't fully checked), then there will in fact be a canonical isomorphism $[X, -] \cong G$ where $[-, -]$ is the internal hom in $M$ (here $[X, -]$ provides the "usual" factorization). Provided am now editing that $G$ is thus enriched, the idea response out as it is first to produce a canonical arrow $[X, -] \to G$; superseded by the enriched Yoneda lemma, this corresponds to an arrow $I \to G(X)$ where $I = F(1)$ is the monoidal unit of $M$; this in turn corresponds to the canonical arrow

$$1 \to \hom_M(X, X) \cong UGX$$

that names the identity on $X$. Then, applying $U$ to $[X, -] \to G$, we get the isomorphism $id: \hom(X, -) \to \hom(X, -)$, but since a monadic functor $U$ reflects isomorphisms, we see that $[X, -] \to G$ already had to be an isomorphism. In other words, up to isomorphism $G$ coincides with the "usual" lift $[X, -]$.

Continuing this train of thoughtMartin's second comment below, suppose $\{\theta_i: X \to \sum_{ar(\theta_i)} X\}$ and $\{\eta_i: X \to \sum_{ar(\eta_i)} X\}$ are two coalgebra structures on $X$ which lift $\hom(X, -)$ to $M$-algebra functors $[X, -], [X, -]': M \to M$ respectively. By the above, we know that $[X, -]$ and $[X, -]'$ are isomorphic; more exactly that there is an algebra-valued isomorphism $[X, -] \to [X, -]'$ lying over makes the identity $\hom(X, -) \to \hom(X, -)$ (and there is just one such transformation since $U$ is faithful)situation quite clear. But this says precisely that the identity on $X$ is a coalgebra map from the one coalgebra structure $\{\theta_i\}$ on $X$ to the other $\{\eta_i\}$, which is just another way of saying Apologies for the two coalgebra structures on $X$ coincidenoise.

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that names the identity on $X$. Then, applying $U$ to $[X, -] \to G$, we get the isomorphism $id: \hom(X, -) \to \hom(X, -)$, but since a monadic functor $U$ reflects isomorphisms, we see that $[X, -] \to G$ already had to be an isomorphism. In other words, up to isomorphism $G$ coincides with the "usual" lift $[X, -]$.

Continuing this train of thought, suppose $\{\theta_i: X \to \sum_{ar(\theta_i)} X\}$ and $\{\eta_i: X \to \sum_{ar(\eta_i)} X\}$ are two coalgebra structures on $X$ which lift $\hom(X, -)$ to $M$-algebra functors $[X, -], [X, -]': M \to M$ respectively. By the above, we know that $[X, -]$ and $[X, -]'$ are isomorphic; more exactly that there is an algebra-valued isomorphism $[X, -] \to [X, -]'$ lying over the identity $\hom(X, -) \to \hom(X, -)$ (and there is just one such transformation since $U$ is faithful). But this says precisely that the identity on $X$ is a coalgebra map from the one coalgebra structure $\{\theta_i\}$ on $X$ to the other $\{\eta_i\}$, which is just another way of saying the two coalgebra structures on $X$ coincide.

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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.)

As for question 2: I haven't nailed down all the details, but if $G$ is the lift of $\hom(X, -): M \to Set$ provided by a co-$T$-algebra structure on $X$ internal to $M$, then provided that the canonical natural map

$$\hom_M(X, Y\hom_M(A, B) \to \hom_M(GX, GY)$$ hom_M(GA, GB)$$is an algebra map (as I think it will be, but these are the details I haven't fully checked), then there will in fact be a canonical isomorphism [X, -] \cong G where [-, -] is the internal hom in M (here [X, -] provides the "usual" factorization). Provided that G is thus enriched, the idea is first to produce a canonical arrow [X, -] \to G; by the enriched Yoneda lemma, this corresponds to an arrow I \to G(X) where I = F(1) is the monoidal unit of M; this in turn corresponds to the canonical arrow$$1 \to \hom_M(X, X) \cong UGX

that names the identity on $X$. Then, applying $U$ to $[X, -] \to G$, we get the isomorphism $id: \hom(X, -) \to \hom(X, -)$, but since a monadic functor $U$ reflects isomorphisms, we see that $[X, -] \to G$ already had to be an isomorphism.

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