Do symmetric monoidal functors preserve reflexivity?

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Suppose that $(C, \otimes_C, \mathcal{H}om_C, I_C), (D, \otimes_D, \mathcal{H}om_D, I_D)$ are closed symmetric monoidal categories. Let $$ F: (C, \otimes_C, I_C) \rightarrow (D, \otimes_D, I_D) $$ be a (strong) symmetric monoidal functor, i.e. $F(X \otimes_C Y) \cong F(X) \otimes_D F(Y)$ and $F(I_C) \cong I_D$. Then $F$ preserves strongly dualizable objects, but in general, it is not true that $F$ preserves internal homs and hence in general $F$ does not preserve weak duals, i.e. $F(X^\vee) \not\cong F(X)^\vee$, where $X^\vee:=\mathcal{H}om(X, I_C)$ is a weak dual.

Next, suppose further that $(C, \otimes_C, \mathcal{H}om_C, I_C)$ is $*$-autonomous, i.e. has a global dualizing object $G$. Namely, if we set $A^\vee: = \mathcal{H}om_C(A, G)$, then the canonical map $$ A \rightarrow (A^\vee)^\vee $$ is an isomorphism.

Question: Is it true that $F(G)$ is a global dualizing object for $(D, \otimes_D, \mathcal{H}om_D, I_D)$, making this category also $*$-autonomous? Alternatively, is the essential image of $F$ $*$-autonomous, i.e. is $$ F(A) \rightarrow (F(A)^\vee)^\vee $$ an isomorphism for all objects $A$ in $C$? Here, $F(A)^\vee:= \mathcal{H}om_D(F(A), F(G))$.

Note I know that certain categories of topological vector spaces are $*$-autonomous with respect to internal hom given by continuous linear maps: https://www.math.mcgill.ca/barr/papers/tvs.pdf On the other hand, the category of (infinite-dimensional) vector spaces is not $*$-autonomous, as it fails the condition $A\rightarrow (A^\vee)^\vee$. I thought that the forgetful functor that takes a topological vector space to the underlying discrete vector space would answer the above question, but in fact this functor is not strongly symmetric monoidal since the tensor product of topological vector spaces is a quotient of the algebraic tensor product. But maybe a similar example works?