A question on the Lawvere theory of vector spaces 
Let $k$ be a field and write $L$ for the Lawvere theory of $k$-vector spaces, which is the category of f.d. $k$-vector spaces.

By definition, models of $L$ in sets is the category of $k$-vector spaces.
What is the category of models in f.d. $k$-vector spaces? Is it $L$ itself?
 A: Chasing through the definitions, a model of $L$ in $\mathbf{Vect}_k$ is ultimately a pair $(V, \rho)$ consisting of a $k$-vector space $V$ and a ring homomorphism $\rho : k \to \operatorname{End}(V)$.
Say that a standard model is one where $\rho(z)$ is scalar multiplication by $z$.
Now, consider $k = \mathbb{C}$ and the model $T = (\mathbb{C}, \rho)$ where $\rho(z) w = \bar{z}w$. Observe that this really is a ring morphism in $z$ and $\mathbb{C}$-linear in $w$.
Theorem: $T$ is not isomorphic to any standard model
Proof: Let $C$ be the standard model on $\mathbb{C}$. A morphism of models is completely determined by the map on underlying vector spaces. Suppose there were a homomorphism $f : T \to C$. Then, for each complex $z$ we have a commutative square
$$ \require{AMScd} \begin{CD}
\mathbb{C} @> \rho(z) >> \mathbb{C}
\\ @VVfV @VVfV
\\ \mathbb{C} @> z >> \mathbb{C}
\end{CD} $$
Plugging in an arbitrary complex $w$ we conclude $ f(\bar{z}w) = z f(w) $, and consequently $ \bar{z} w f(1) = z w f(1) $.
Thus, we conclude that $f(1) = 0$. There does not exist any isomorphism $T \to C$. $\square$
A: This answer is to supplement the answer already given by Hurkyl. 
It's true in this case that the Lawvere theory $L$ is equivalent to the category $\text{Vect}_{fd}$ of f.d. vector spaces (morally speaking it's opposite to this category -- the Lawvere theory is on general grounds opposite to the category of f.g. free algebras -- but here this category happens to be equivalent to its opposite via linear duality, so we're safe). The problem is to characterize the category of functors $L \to \text{Vect}_{fd}$ that preserve finite products. Or what is the same, functors $\text{Vect}_{fd} \to \text{Vect}_{fd}$ that preserve finite products, or such functors that preserve finite coproducts -- by linear duality there are various ways of thinking about it. 
Now $\text{Vect}_{fd}$ is $\text{Ab}$-enriched, and coproduct- or biproduct-preserving functors $\text{Vect}_{fd} \to \text{Vect}_{fd}$ are the same as $\text{Ab}$-enriched functors. Moreover, $\text{Vect}_{fd}$ is, qua the $\text{Ab}$-enriched world,  the Cauchy completion of $k$ viewed as a one-object $\text{Ab}$-enriched category (the more general statement is that the $\text{Ab}$-enriched Cauchy completion of a ring $R$ is the category of f.g. projective $R^{op}$-modules). It follows that coproduct-preserving functors $\text{Vect}_{fd} \to \text{Vect}_{fd}$ are equivalent to $\text{Ab}$-enriched functors $k \to \text{Vect}_{fd}$. These are given by a finite-dimensional vector space $W$ and a ring homomorphism 
$$k \to \text{Vect}_k(W, W).$$ 
In slightly different language: this data can be described as a $(k, k)$-bimodule (a left $k$- right $k$-bimodule) $W$ whose left $k$-module structure is finite-dimensional. In that case, viewing $\text{Vect}_{fd}$ as a category of left $k$-modules, we get a (co)product-preserving endofunctor as 
$$W \otimes_k -: \text{Vect}_{fd} \to \text{Vect}_{fd}$$ 
and all product-preserving endofunctors are of this form. 
It is clear that any finite-dimensional vector space $W$ gives such a bimodule (just pull back the $k$-scalar action along the map $k \otimes k \to k$ given by $k$-multiplication), but as Hurkyl's argument shows, these do not exhaust all such bimodules. 
