Checking when a linear isomorphism is natural

The idea that a map is 'natural' is sometimes informally stated as being independent of choice of basis, but what it really means is that it's a component of a natural transformation between two functors.

Here, the functors are $\operatorname{Hom}(-\otimes \mathbb{C},\mathbb{R})$ and $\operatorname{Hom}(-,\mathbb{R})\otimes\mathbb{C}$. The question of whether they're naturally isomorphic requires you to produce a natural isomorphism between these two functors, which means a choice of isomorphism $$\alpha_V:\operatorname{Hom}(V\otimes \mathbb{C},\mathbb{R})\to\operatorname{Hom}(V,\mathbb{R})\otimes\mathbb{C}$$ for every choice of vector space $V$ such that for any map $f:W\to V$, you obtain a commutative square

$$\begin{matrix}&\alpha_W&\\\operatorname{Hom}(W\otimes \mathbb{C},\mathbb{R})&\cong&\operatorname{Hom}(W,\mathbb{R})\otimes\mathbb{C}\\\downarrow &&\downarrow\\\operatorname{Hom}(V\otimes \mathbb{C},\mathbb{R})&\cong&\operatorname{Hom}(V,\mathbb{R})\otimes \mathbb{C}\\&\alpha_V& \end{matrix}$$

These two functors may be naturally isomorphic, but, at least to me, not in an obvious way.