Let $V$ be an infinite-dimensional vector space having a countable basis, let $V'$ be its dual space and let $R: V' \rightarrow V'$ be a linear map. Is it always true that there is a map $T: V \rightarrow V$ such that $R$ is its dual map, i.e. $R = T'$?
So the answer makes sense to be "no". As I do not understand cardinals, it would be great if one can give an explicit counterexample. Any help appreciated!
EDIT: both $\aleph_0^{\aleph_0}<\aleph_1^{\aleph_1}$ and $\aleph_0^{\aleph_0}=\aleph_1^{\aleph_1}$ are consistent with the ZFC axioms (note that the former is forced by the Continuum Hypothesis). This and the argument below show at the very least that the answer to your question cannot be a solid yes. See Asaf Karagila's answer here for more insight about $(1)$.
My goal is to prove that as soon as $V$ is "big enough", this is impossible for cardinality reasons as $|\operatorname{End}V|<|\operatorname{End}V^*|$ (so there can be no surjective map from $\operatorname{End}V$ to $\operatorname{End}V^*$).
I will need the following result for which I couldn't find a reference but which seems reasonable: $$\text{If $a,b$ are cardinals such that $a<b$, then $a^a<b^b$.}\tag{1}$$
Since an endomorphism is determined by the choice of an element of $V$ for each element of a basis, you have $$|\operatorname{End}V|=|V|^{\dim V}\qquad\text{ and }\qquad|\operatorname{End}V^*|=|V^*|^{\dim V^*}$$
Lemma$^{\star}$: If $|V|>|k|$ where $k$ is the ground field, then $\dim V=|V|$.
Therefore, assuming $|V|>|k|$, you have $$|\operatorname{End}V|<|\operatorname{End}V^*|\Longleftrightarrow |V|^{|V|}<|V^*|^{|V^*|}$$
Lemma$^{\star}$: If $V$ is not finite-dimensional, then $\dim V<\dim V^*$.
Corollary (of the two lemmas together): If $|V|>|k|$, then $|V|<|V^*|$.
Conclusion (up to $(1)$):$$|V|>|k|\implies |\operatorname{End}V|<|\operatorname{End}V^*|$$
$^{\star}$(note that this is a link)