It is known that a principally polarized complex torus $T$ is isomorphic to its dual. Indeed, if $L$ is a line bundle over $T$, s.t. $c_1(E)$ is a principal polarization on $T$, then the mapping $x \rightarrow \tau_x^*(L) \otimes L^{-1}$ is an isomorphism from $T$ to $\hat{T}$ (here $\tau_x$ is the shift by $x$, and $^*$ denotes the inverse image). However, I could not find any suggestion in the Internet in regard of some sort of a converse statement. That is, if $T$ is isomorphic to its dual, is it possible to derive some properties of $T$. Maybe the existence of a principle polarization is the necessary and sufficient condition.
I suppose that the answer could be simpler under the condition of $T$ being algebraic, so there exists some polarization on $T$, not necessarily a principle one. Is it possible to prove that an algebraic torus which is isomorphic to its dual admits a principle polarization? I would be content with an answer to this simpler case, but it is also interesting whether the situation is similar in the non-algebraic case.
This is not a complete answer, but too long for a comment.
Birkenhake and Lange[1] give the following criterion which indicates that not any morphism $X \to \hat X$ is induced by a line bundle:
Here $\overline \Omega = \operatorname{Hom}_{\overline{\mathbb{C}}}(V, \mathbb C)$ is the vector space of $\mathbb C$-antilinear forms $l\colon V \to \mathbb C$, and $\hat X = \overline\Omega / \hat \Lambda$ for the lattice $$\hat \Lambda = \{\, l \in \overline \Omega:l(\Lambda) \subset \mathbb Z \,\}.$$ Maybe try to construct a counterexample by giving an $F$ which does not induce a hermitian form?
[1] Birkenhake, Lange, Complex Abelian Varieties