Is the identity functor naturally isomorphic to a covariant dual functor?

Suppose we had a natural transformation $\eta: 1 \Rightarrow D$. Let $V$ be a vector space and $T \in GL(V)$. The naturality condition asserts that for all $v \in V$, we have $$\eta_V(Tv) = \eta_V(v) \circ T^{-1}.$$ Fix $v \in V \setminus \{0\}$, and let $f = \eta_V(v)$. Since $GL(V)$ acts transitively on $V \setminus \{0\}$, we see that $\eta_V: V \to V^*$ is completely determined by $f$ according to the equation above.

We need to check that $\eta_V$ is thus well-defined. In particular, we need to have $$f = f \circ T$$ for all $T$ fixing $v$.

Let us be more concrete. Let $V = \mathbb{R} \langle e_1, e_2, e_3 \rangle$ and $v = e_1$. Consider transformations $T \in GL(V)$ as follows: $$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$ These matrices all fix $e_1$. Plugging them into the equation $f = f \circ T$, we find \begin{align} f(e_2) &= f(e_1) + f(e_2) \\ f(e_3) &= f(e_2) + f(e_3) \\ f(e_2) &= f(e_3) \end{align}

Consequently, we have $f(e_1) = f(e_2) = f(e_3) = 0$, so that $f$ is the zero linear functional. But then $\eta_V(e_1) = 0 = \eta_V(0)$, contradicting the fact that $\eta_V$ is a linear isomorphism. So no natural transformation $\eta: 1 \Rightarrow D$ can exist.