A non-projective rigid object in an abelian monoidal category

What is an example of a rigid object $$A$$ in an abelian monoidal category $$\mathcal{M}$$ that is not projective as an object in $$\mathcal{M}$$? (Since $$\mathcal{M}$$ is abelian projective just means that all exact sequences of the form $$0 \to R \to Q \to P \to 0$$ split.) What are sufficient conditions for such a rigid object to be projective?

If by "rigid" you mean the same thing as dualizable, an example is given by $$\mathcal M = Mod_{R[G]}$$ for some commutative ring $$R$$ and group $$G$$, with monoidal structure given by $$\otimes_R$$ and the diagonal action.
Then the object $$R$$ with a trivial $$G$$-action is the unit, hence it's rigid, but it's rarely projective: $$\hom(R,-)$$ is isomorphic to $$(-)^G$$, so it's projective if and only if the latter is exact, if and only if $$H^*(G,-)$$ vanishes on $$R$$-modules - counterexamples are given by $$R= \mathbb F_p, G=$$ any finite group with order divisible by $$p$$.
More generally, note the following : if $$P$$ is dualizable, with dual $$P^\vee$$ then $$\hom(P,-)\cong \hom(\mathbf 1, P^\vee \otimes -)$$, where $$\mathbf 1$$ is the unit. So there are two ways exactness can fail: $$\hom(\mathbf 1,-)$$ might not be exact (in which case, the unit is an example anyway), or $$P^\vee\otimes -$$ might not be exact (these are not necessary/sufficient conditions, because $$\hom(\mathbf 1,-)$$ might not be conservative either).
Note that in fact, the latter cannot happen if $$\mathcal M$$ is symmetric monoidal : indeed $$P^\vee\otimes -$$ is then both a left adjoint (hence right exact) and a right adjoint (hence left exact). I don't know what happens if $$\mathcal M$$ is just monoidal, because then $$P^\vee$$ might itself not be dualizable (say I fixed a side and am only referring to "left dualizable")
So in deciding if $$P$$ is projective, you really only have to check whether the unit $$\mathbf 1$$ is projective.
Note that it might still happen that the unit is not projective, yet $$P$$ is. Consider for instance the following : take a counterexample $$\mathcal M$$ as above, and an ordinary category of modules $$\mathcal M'$$ over a commutative ring. Then the unit of $$\mathcal M\times\mathcal M'$$ is not projective (it has the unit of $$\mathcal M$$ as a summand, and that one is not projective), yet any projective module in $$\mathcal M'$$ is projective in the product.