If a $\otimes$-idempotent object has a dual, must it be self-dual?

If $C$ is not assumed to be symmetric, then the answer to questions 1 and 2 is no. Let $p^* \dashv p_* \dashv p^!$ be a fully faithful adjoint triple with $p_* : A \to B$. (For instance, $A= \mathrm{Top}$ and $B=\mathrm{Set}$ with $p_*$ the forgetful functor, $p^*$ the discrete topology, and $p^!$ the indiscrete one.) Then $F = p^* p_*$ is an idempotent comonad on $A$ and $U = p^! p_*$ is an idempotent monad, and $F \dashv U$. Thus, $F$ is well-idempotent and dualizable in the endofunctor category $C = [A,A]^{\mathrm{op}}$ with its composition monoidal structure, but is not generally self-dual.

EDIT: It looks like I was too hasty to delete this answer. I think I have fixed the gap that Anton noted in the comments and in his answer. As Anton pointed out in his answer, the trace loop that I dropped out at the end is nontrivial, but it can be removed from any diagram where an "X" appears in another connected component. Intuitively, an idempotent should have dimension 1 wherever it is nonzero. I've oriented the string diagrams vertically so that any horizontal cross-section can be read left-to-right in symbolic order.

The answer to question 2 (hence 3) is "yes" for symmetric monoidal categories. Let $f:X\otimes X\to X$ be the two-sided inverse of $id_x\otimes i$. As it turns out, $f$ is also the left inverse of $i\otimes id_X$. This is proven in the image linked below.

Let $\phi:X\to X^\vee$ be the composition

$X\to X\otimes I\to X\otimes X\otimes X^\vee\to X\otimes X^\vee\to I\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X^\vee\to X^\vee\otimes I\to X^\vee,$

and let $\psi:X^\vee\to X$ be the composition $X^\vee\to I\otimes X^\vee\to X\otimes X^\vee\to I\to X$. These two maps are mutually inverse. Here is a string diagram "proof" of both of the above facts.

This is the old proof with an unjustified step at the end.

Ok, I have no idea how to fix the invariant of string diagrams, but I have an explicit counterexample. Consider the 2-category of distributors $Dist$: its objects are small categories, its morphisms from $C$ to $D$ are bifunctors $C\times D^{op} \to Set$, its 2-morphisms are natural transformations and the composition is given by tensor product of bifunctors. It is a symmetric monoidal 2-category with monoidal product given by $\times$. Its equivalences are Morita equivalences of categories, which are the same as colimit-preserving equivalences of their presheaf categories. Any category is a dualizable object in $Dist$, the dual is given by the opposite category and the evaluation-coevaluation morphisms are given by hom-functors. We can extract a symmetric monoidal 1-category from it by discarding all noninvertible 2-morphisms and collapsing all invertible ones, i.e. it is the 1-truncation of the 1-categorical core. A simpler and similar example would be the 2-category of associative algebras with bimodule categories for morphisms, but I can't produce a counterexample in it.

Anyway, for any category $C$ its infinite cartesian power $C^{\mathbb N}$ is obviously idempotent. It's also a category of functors from $\mathbb N$, so all its morphisms and limits/colimits are pointwise. Consider now the category of nonempty finite sets $Fin_{ne}$. I claim that $\left(Fin_{ne}\right)^{\mathbb N}$ and $\left(Fin^{op}_{ne}\right)^{\mathbb N}$ are not Morita equivalent. They are certainly not equivalent as categories since $Fin_{ne}$ (and thus $\left(Fin_{ne}\right)^{\mathbb N}$) has a final object and no initial object while for $Fin^{op}_{ne}$ it is the other way round. But $Fin_{ne}$ (and thus $Fin^{op}_{ne}$) are Cauchy complete (all idempotents split), so the same is true for $\left(Fin_{ne}\right)^{\mathbb N}$ and $\left(Fin^{op}_{ne}\right)^{\mathbb N}$. Any Morita equivalence between Cauchy complete categories is realised as an actual equivalence of categories, by Theorem 2.7.3, QED.