In a rigid category every morphism of monoidal functors is an isomorphism

Because $\mathcal{C}$ is rigid, choose for each object $X$ a left dual $X^{\vee }$. Since $F$ and $G$ are strong monoidal, they send the duality maps of $X$ to duality maps of $F(X)$ and $G(X)$: after transporting the images of ${\mathrm{ev}}_X$ and ${\mathrm{coev}}_X$ across the strong monoidal structure isomorphisms, $F(X^{\vee })$ is a left dual of $F(X)$ and $G(X^{\vee })$ is a left dual of $G(X)$.

Define ${\xi }_X:G(X)\to F(X)$ using the transported duality maps by the composite $G(X)\cong \mathbf{1}\otimes G(X)\stackrel{{\mathrm{coev}}_{F(X)}\otimes 1}{\to }F(X)\otimes F(X^{\vee })\otimes G(X)\stackrel{1\otimes {\eta }_{X^{\vee }}\otimes 1}{\to }F(X)\otimes G(X^{\vee })\otimes G(X)\stackrel{1\otimes {\mathrm{ev}}_{G(X)}}{\to }F(X)\otimes \mathbf{1}\cong F(X)\mathrm{.}$ This construction uses $F(X^{\vee })$ and $G(X^{\vee })$ as left duals of $F(X)$ and $G(X)$ from step 1.1; it does not identify a left dual of $F(X^{\vee })$ with $F(X)$.

Expand ${\xi }_X{\eta }_X$ using step 2.1. Naturality of $\eta$, together with its tensor and unit compatibility from [L2], moves ${\eta }_X$ across the coevaluation and evaluation; the remaining composite is the zig-zag identity for the dual pair $F(X^{\vee }),F(X)$. Hence ${\xi }_X{\eta }_X=1_{F(X)}$. The mirrored calculation uses the zig-zag identity for $G(X^{\vee }),G(X)$ and gives ${\eta }_X{\xi }_X=1_{G(X)}$. Thus every component ${\eta }_X$ is an isomorphism, so $\eta$ is a natural isomorphism.