Owen Biesel , that doesn't seem right actually.

You seem to be taking a scheme, which is a functor \$$F: \mathcal{C} \to \mathbf{Set}\$$, and then applying it *backwards* along a functor \$$G\$$ mapping between the categories \$$\mathcal{D} \to \mathcal{C}\$$.

If intuition serves, \$$G\$$ should mapping both tables into the single table. That is to say, \$$G\$$ is an inclusion of sorts.

Therefor, the instance \$$(F \circ G): \mathcal{D} \to \mathbf{Set}\$$ should be one big table,

\$\begin{array}{c|c} \text{People} & \mathrm{FriendOf} \\\\ \hline Alice & Bob \\\\ Bob & Alice \\\\ \vdots & \vdots \\\\ Stan & Tyler \\\\ Tyler & Stan. \\\\ \vdots & \vdots \\\\ Adele & Sara\\\\ Bertram & Antonio \\\\ \vdots & \vdots \\\\ Siegmund & Teresa \\\\ \vdots & \vdots \\\\ Antonio & Bertram \\\\ Bruno & Bertram \\\\ \vdots & \vdots \\\\ Sara & Adele \\\\ Teresa & Siegmund. \\\\ \vdots & \vdots \end{array} \$

Edit: Note that \$$Tyler\$$ doesn't show up twice and that his Italian friend gets mapped to someone else as a friend.