Can the underlying set functor corresponding to an algebraic theory always be viewed as a model of that theory? Let $\mathsf{T}$ denote a Lawvere theory, and let $\mathbf{C}$ denote its category of models in $\mathbf{Set}$. Write $U : \mathbf{C} \rightarrow \mathbf{Set}$ for the underlying set functor. I think that $U$ can always be viewed as model of $\mathsf{T}$ in the functor category $[\mathbf{C},\mathbf{Set}].$ For example, suppose $\mathsf{T}$ has a binary operation $f : G \times G \rightarrow G$, where $G$ is the generic object of $\mathsf{T}$. Then there should be a corresponding natural transformation $\nu : U \times U \Rightarrow U$ given by writing $\nu_X = X(f)$ for all objects $X$ of $\mathbf{C}$. Hence $U$ becomes equipped with the operations of $\mathsf{T}$ in a natural way.
Is this right? If so, I'd appreciate some kind of a discussion of how this all works. Its currently very fuzzy to me.

In more detail: Recall that a model of a Lawvere theory $\mathcal{T}$ in a category (with finite products) $\mathcal{S}$ is a finite-product-preserving functor $\mathcal{T} \to \mathcal{S}$. Thus, for any category $\mathcal{C}$, a $\mathcal{T}$-model in $[\mathcal{C}, \mathbf{Set}]$ is the same thing as a diagram $\mathcal{C} \to \mathbf{Mod}(\mathcal{T}, \mathbf{Set})$; in symbols:
$$\mathbf{Mod}(\mathcal{T}, [\mathcal{C}, \mathbf{Set}]) \cong [\mathcal{C}, \mathbf{Mod}(\mathcal{T}, \mathbf{Set})]$$
In particular, we may take $\mathcal{C} = \mathbf{Mod} (\mathcal{T}, \mathbf{Set})$. Then,
$$\mathbf{Mod}(\mathcal{T}, [\mathbf{Mod} (\mathcal{T}, \mathbf{Set}), \mathbf{Set}]) \cong [\mathbf{Mod} (\mathcal{T}, \mathbf{Set}), \mathbf{Mod}(\mathcal{T}, \mathbf{Set})]$$
and the $\mathcal{T}$-model structure on the forgetful functor $\mathbf{Mod} (\mathcal{T}, \mathbf{Set}) \to \mathbf{Set}$ is precisely the one corresponding to the identity functor on $\mathbf{Mod} (\mathcal{T}, \mathbf{Set})$.
In fact, all of this goes through for any category $\mathcal{S}$ with finite products instead of $\mathbf{Set}$.
