The result does not require $\mathcal{C}'$ to be rigid. But of course a monoidal functor with rigid domain will have rigid image, so any nonrigidity in $\mathcal{C}'$ is not seen by the problem.
Anyway, the structure of the argument goes:
- Suppose $F : \mathcal{C} \to \mathcal{C}'$ is a (strong) monoidal functor, and $X \in \mathcal{C}$ is a dualizable object. Then $F(X)$ is dualizable, with duality data given by applying $F$ to the duality data for $X$. Idea of proof: suppose $(X^\vee, \mathrm{pairing}, \mathrm{copairing})$ are duality data for $X$. This means that they have prescribed domain and codomain, and importantly that a certain equation (called "zig-zag" or "zorro" or "snake" or "triangle" or probably other things depending on what school you come from) holds. But functors preserve equations.
- Suppose $\lambda : F \Rightarrow G$ is a monoidal natural transformation, and $X$ is a dualizable object with dual $X^\vee$. Then $\lambda_X : FX \to GX$ is invertible, with inverse the transpose of $\lambda_{X^\vee}$. Let's type-check this: $\lambda_{X^\vee} : F(X^\vee) \to G(X^\vee)$, but $F(X^\vee) \cong (FX)^\vee$, and this isomorphism is canonical by step 1 together with the uniqueness of dual objects. Idea of proof: It suffices to prove that $\mathrm{pairing}_{GX}(\lambda_X \otimes \lambda_{X^\vee}) = \mathrm{pairing}_{FX}$, and similarly for $\mathrm{copairing}$. But this is immediate from naturality and monoidality of $\lambda$.
- This suffices: if a monoidal natural transformation takes invertible values on objects, then its is monoidally invertible. Idea of proof: It is standard that if a natural transformation takes invertible values, then it is invertible — the main thing to show that the inverse values are also natural, but this is easy. The other thing is to show that the inverse transformation is monoidal; again this is essentially straightforward.
