An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $\langle T^*Tx,x\rangle=\langle x,x\rangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $\ell^2$, the operator sending $(a_0,a_1,a_2,a_3,\ldots)$ to $(0,a_0,a_1,a_2,\ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).