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We cannot speak of a natural isomorphism directly between $F$ and $G$ because natural isomorphisms compare functors with the same domain and codomain. Instead, the video reformulates the condition: $F$ and $G$ are adjoint if and only if the two curried hom-functors $\mathcal{B}(F(-),-)$ and $\mathcal{A}(-,G(-))$, both mapping from $\mathcal{A}^{op} \times \mathcal{B}$ to SET, are naturally isomorphic.

Conditions: $F:\mathcal{A}\to\mathcal{B}$ and $G:\mathcal{B}\to\mathcal{A}$; Both sides are regarded as functors out of $\mathcal{A}^{op}\times\mathcal{B}$; The natural isomorphism is between functors with the same domain and codomain

We cannot speak of a natural isomorphism directly between $F$ and $G$ because natural isomorphisms compare functors with the same domain and codomain. Instead, the video reformulates the condition: $F$ and $G$ are adjoint if and only if the two curried hom-functors $\mathcal{B}(F(-),-)$ and $\mathcal{A}(-,G(-))$, both mapping from $\mathcal{A}^{op} \times \mathcal{B}$ to SET, are naturally isomorphic.

Conditions: $F:\mathcal{A}\to\mathcal{B}$ and $G:\mathcal{B}\to\mathcal{A}$; Both sides are regarded as functors out of $\mathcal{A}^{op}\times\mathcal{B}$; The natural isomorphism is between functors with the same domain and codomain