@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.