Collision attack on hash
It's straightforward to find a large set of preimages for any hash value!

For arbitrary $x,z \in \{0,1\}^n$, you can construct $Y(x,z)$ such that $h(x||Y(x,y))=z$. Let $D$ be the decryption function for the block cipher, i.e. $D(x,\cdot)$ is the inverse permutation from $E(x,\cdot)$. The goal $h(x||Y(x,z))=z$, i.e. $E(x,Y(x,z))=z$, is equivalent to $Y(x,z)=D(x,z)$.

Thus for any $x$, all of the inputs $x||D(x,z)$ for $z \in \{0,1\}^n$ are collisions for $h$.