@Christopher – yeah, I was thinking along those lines.

So we have \$$\mathcal{X}\$$ and \$$\mathcal{Y}\$$ as \$$\mathcal{V}\$$-categories.

And we have \$$F: \mathcal{X}\to \mathcal{Y}\$$ and \$$G: \mathcal{Y}\to \mathcal{X}\$$ as \$$\mathcal{V}\$$-functors.

Then define \$$\Phi, \Psi : \mathcal{X}^\text{op}\times\mathcal{Y}\to\mathcal{V}\$$ by

$\Phi(x, y) = \mathcal{Y}(F(x), y) \qquad \Psi(x, y) = \mathcal{X}(x, G(y))$

We need to check these are a legit \$$\mathcal{V}\$$-functors.

$(\mathcal{X}^\text{op}\times\mathcal{Y})((x, y), (x', y')) = \mathcal{X}^\text{op}(x, x')\otimes\mathcal{Y}(y, y') = \mathcal{X}(x', x)\otimes\mathcal{Y}(y, y')$

$\mathcal{V}(\Phi(x, y), \Phi(x', y')) = \mathcal{V}(\mathcal{Y}(F(x), y), \mathcal{Y}(F(x'), y')) = \mathcal{Y}(F(x), y) \multimap \mathcal{Y}(F(x'), y')$

To show that the top line \$$\leq\$$ the bottom line we need to prove

$\mathcal{Y}(F(x), y) \otimes \mathcal{X}(x', x)\otimes\mathcal{Y}(y, y')\leq \mathcal{Y}(F(x'), y')$

But we know \$$\mathcal{X}(x', x)\leq\\mathcal{Y}(F(x'), F(x))\$$ (because \$$F\$$ is a \$$\mathcal{V}\$$-functor), so

$\mathcal{Y}(F(x), y) \otimes \mathcal{X}(x', x)\otimes\mathcal{Y}(y, y')\leq \mathcal{Y}(F(x), y) \otimes \mathcal{Y}(F(x'), F(x))\otimes\mathcal{Y}(y, y') \cong \mathcal{Y}(F(x'), F(x))\otimes\mathcal{Y}(F(x), y) \otimes\mathcal{Y}(y, y')$

which is \$$\leq \mathcal{Y}(F(x'), y')\$$ by the "composition" rule of \$$\mathcal{Y}\$$ as \$$\mathcal{V}\$$-category (applied twice).

I'm guessing the \$$\Psi\$$ case works similarly, and then \$$\Phi \cong \Psi \iff F \dashv G\$$ is obvious.