A quine in pure lambda calculus
One one hand this is impossible, because a quine is supposed to output its own code, and the pure lambda calculus has no means for performing output.

On the other hand, if you assume that the resulting term is the output, then every normal form is a quine.

For example, lambda term $(\lambda x. x)$ is already a normal form, then assuming that its output is the resulting normal form, the output is $(\lambda x. x)$. Thus $(\lambda x. x)$ is a quine.