At the end of the LMN 137, "Reports of the Midwest Category Seminar IV", there is a list of exercises.
"5. Considering a left-adjoint as male and a right adjoint as female, give the correct term for a contravariant functor self-adjoint on the right."
I am not sure I understand how to solve (or get the pun) of the exercise.
What would be a contravariant functor that is a self-adjoint on the right?
I have the following in mind: $F : C^{\mathrm{op}} \to C$ such that
$$ C(X,F Y) \cong C(Y, F X) $$
is this right? Is that the definition of a self-adjoint functor on the right?
If $T\colon A\to B$ is a contravariant functor, let denote by $\overline{T}\colon A^{op}\to B$ the corresponding covariant functor.
Definition(Freyd): Two conravariant functors $T\colon A\to B$ and $S\colon B\to A$ are adjoint on the right, if the functor $\overline{S}\colon B^{op}\to A$ is right adjoint(in usual sense) to $\overline{T}^{op}\colon A\to B^{op}$.
So, if $F\colon C\to C$ is a contavariant endofuctor, then it is self-adjoint on the right, if the functor $\overline{F}\colon C^{op}\to C$ is right adjoint(in usual sense) to $\overline{F}^{op}\colon C\to C^{op}$, or, equivalently, if there exists a natural bijection: $$ hom_{C}(X,F(Y))\cong hom_C(Y, F(X)), $$ so you are right.
Well, if we consider the very very strange interpretation from your exercise, then $F$ is a female, whose dual is male...