From Anton Fetisov's work I think we can extract a simple example. In fact this example should be universal for the class of examples he constructs.


Recall that a connected second-countable one-manifold with boundary is a closed interval, an open interval, a half-open interval, or a circle. 

Consider the category whose objects are pairs of countable sets $(A^+,A^-)$ and where a morphism $(A^+, A^-) \to (B^+, B^-)$ is a second-countable 1-manifold with boundary, whose boundary equals $A^+ \cup A^- \cup B^+ \cup B^-$, and such that each closed interval component has one boundary point in $A^+ \cup B^-$ and one in $A^- \cup B^+$ (up to isomorphism).

Composition of a morphism $(A^+ , A^-) \to (B^+, B^-)$ with a morphism $(B^+,B^-) \to (C^+,C^-)$ is obtained by gluing the manifolds along $B$.

A symmetric monoidal structure is given by disjoint union of sets and morphisms. The identity is a union of closed intervals connecting each element to itself.

If I did this with finite sets and compact manifolds, so that the only components were closed intervals and circles, I would have constructed the free symmetric monoidal category on one generator and its dual. Because I allow infinite sets, I must necessarily allow connected components that are not closed.

The element $(\mathbb N,\emptyset)$ is idempotent as $(\mathbb N,\emptyset) \otimes (\mathbb N,\emptyset)= (\mathbb N \cup \mathbb N,\emptyset \cup \emptyset) = (\mathbb N , \emptyset)$.

It has dual $(\emptyset, \mathbb N)$, with the unit and counit each given by a union of closed intervals connecting the two copies of $\mathbb N$, where the relations hold by an infinite disjoint union of the usual stringy proof.

However, it is not self dual, as $(\mathbb N,\emptyset)$ and $(\emptyset,\mathbb N)$ are not isomorphic, as every map between them has all elements connected to half-open intervals (because $B^- \cup A^+$ is empty), so the composition also has all elements connected to half-open intervals and thus isn't the identity.