Is a functor which has a left adjoint which is also its right adjoint an equivalence ?

There are lots of examples. Here's what I think is in some sense the minimal one.

Let $C$ be the terminal category $\mathbf{1}$ (one object, and only the identity arrow). Then for any category $D$, a left adjoint to the unique functor $G: D \to \mathbf{1}$ is an initial object of $D$, and a right adjoint is a terminal object. So, we're looking for a category $D$ that has a zero object (one that is both initial and terminal), but is not equivalent to the terminal category.

There are plenty of such categories $D$, e.g. $\mathbf{Vect}$. But I guess the minimal one is the category $D$ generated by a split epimorphism. In other words, it consists of two objects, $0$ and $d$, and non-identity arrows $$ p: d \to 0, \ \ \ i: 0 \to d, \ \ \ ip: d \to d, $$ satisfying $pi = 1_0$. Then $0$ is a zero object but $D$ is not equivalent to the terminal category.

Yes, there are many such functors. They are usually called "biadjoint." A good example is tensor product with a vector space $V$ in the category of finite dimensional vector spaces. This is actually adjoint to itself.

This is a little funny since to find this adjunction you have to pick an isomorphism $V\cong V^*$, but that's OK; adjunction of functors only makes sense up to isomorphism anyways.

Another good example is induction and restriction for an inclusion of finite groups.

The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is a left adjoint to F ; and so, is a also a right adjoint to F. This is clear because for all A-modules N, M, one has Hom_A(0,N)=Hom_A(M,0). But, F is not an equivalence.