Closure of regularity under the action of replacing identical pairs of letters
You can do it using homeomorphism. Define $h:\Sigma\rightarrow \Sigma^*$ by: $h(\sigma)=\sigma\sigma$.
Then, $h$ induces a homeomorphism $g$ that is defined by $g(\sigma_1,\dots,\sigma_n)=h(\sigma_1)\dots h(\sigma_n)$.
Then, its easy to see that $shrink(L)=g^{-1}(L)$ (prove this!). From closure properties of regular languages we know that $g^{-1}(L)$ is regular, hence also $shrink(L)$.