If $u=\frac{1+\sqrt5}{2}$, then $u^3=2+\sqrt5$, but $u^2=\frac{3+\sqrt5}{2}$. What is the group that measures the power that makes units look nice?

For $A=\mathbb{Z}[x]/(f)$ with quotient field $K$ and ring of integers $B$, does $U(B)/U(A)$ have a name?

For instance $u = \tfrac{1+\sqrt{5}}{2}$ is a unit in $\mathbb{Q}[\sqrt{5}]$, but neither $u$ nor $u^2$ has integer coefficients in the basis $\{ 1, \sqrt{5} \}$. Of course $u^3$ has integer coefficients (spooky if you haven't tried it!) and in fact $u^n$ has integer coefficients iff $0 \equiv n \mod 3$.

For quadratic fields with basis $\{ 1, \sqrt{n} \}$ for $n$ square-free, one almost always has $U(A) = U(B)$. If not, then $[ U(B) : U(A) ] = 3$.

That's crazy, and it should have a name. For instance, I'd like to find out if the following is true, but I don't even know what to look for:

Is $U(B)/U(A)$ always finite? [ where $B$ is the ring of integers of an order $A$ in a number ring ]

In general, if $A \subset B$ is an extension of rings, I would call $B^{\times}/A^{\times}$ the relative unit group. (I am probably not the only one, but I couldn't say how widespread this is.) I feel reasonably confident that there is no specialized terminology in the case of nonmaximal orders.

To answer your non-terminological question: yes, if $A \subset B$ are orders in the same number field, the group $B^{\times}/A^{\times}$ is finite. This follows from the fact that one can prove the Dirichlet Unit Theorem equally well for a nonmaximal order $A$ in a number field $K$: $A^{\times}$ is still finitely generated with torsion subgroup equal to the roots of unity in $A$ and free rank equal to $r+s-1$, where $r$ is the number of real places and $s$ is the number of complex places of $K$. Thus we have finitely generated abelian groups $A^{\times} \subset B^{\times}$ with the same free rank, so $B^{\times}/A^{\times}$ is finite.