= Solution
If $L\in\operatorname{Pic}^0(X)$, part (iii) makes $\Lambda(L)$ trivial. Pulling it back along $(f,g):Y\to X\times X$ gives
$$
\boxed{(f+g)^*L\simeq f^*L\otimes g^*L.}
$$
Taking $Y=X$, $f=\operatorname{id}_X$, and $g=i$, where $i$ is inversion, gives
$$
\mathcal O_X\simeq L\otimes i^*L,
\qquad
\boxed{i^*L\simeq L^\vee.}
$$
Induction with $f=[n]$ and $g=\operatorname{id}_X$ proves $[n]^*L\simeq L^{\otimes n}$ for $n\ge0$; combining this with inversion proves
$$
\boxed{[n]^*L\simeq L^{\otimes n}\quad(n\in\mathbb Z).}
$$
Conversely, suppose $i^*L\simeq L^\vee$. For $M=\phi_L(x)$, part (ii) gives $M\in\operatorname{Pic}^0(X)$, so the result just proved yields $i^*M\simeq M^\vee$. On the other hand,
$$
\phi_{i^*L}(x)=i^*\phi_L(-x)=i^*(M^\vee)\simeq M,
$$
whereas $i^*L\simeq L^\vee$ gives $\phi_{i^*L}(x)=\phi_{L^\vee}(x)=M^\vee$. Hence $M^{\otimes2}$ is trivial for every $x$, so $\phi_{L^{\otimes2}}$ is trivial and $L^{\otimes2}\in\operatorname{Pic}^0(X)$. The torsion-freeness proved in part (ii) now implies
$$
\boxed{L\in\operatorname{Pic}^0(X).}
$$
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