Here is an explicit realization of the counterexample suggested by Dustin. The field ${\mathbf Q}(\sqrt{10})$ has class number 2 and its Hilbert class field is obtained by adjoining $\sqrt{2}$. The ring of integers ${\mathbf Z}[\sqrt{10}]$ has (fundamental) unit $u:=3+\sqrt{10}$, whose minimal polynomial over ${\mathbf Q}$ is $T^2 - 6T - 1$. The two ideal classes in ${\mathbf Z}[\sqrt{10}]$ are represented by the ideals $(1)$ and $(2,\sqrt{10})$, which have ${\mathbf Z}$-bases $\{1,u\}$ and $\{2,\sqrt{10}\}$. Multiplication by $u$ on these two ideals is represented, using the indicated $\mathbf Z$-bases, by the respective matrices $A = (\begin{smallmatrix}0&1\\1&6\end{smallmatrix})$ and $B = (\begin{smallmatrix}3&5\\2&3\end{smallmatrix})$. These matrices are both in ${\rm GL}_2({\mathbf Z})$, they are not conjugate in this group, but they are conjugate by the matrix $U = (\begin{smallmatrix}\sqrt{2}&5+3\sqrt{2}\\1&3+2\sqrt{2}\end{smallmatrix})$, which lies in ${\rm GL}_2({\mathbf Z}[\sqrt{2}])$. That is, $UAU^{-1} = B$. This conjugating matrix $U$ has determinant $-1$. A matrix with determinant 1 and algebraic integer entries that satisfies $VAV^{-1} = B$ is
$V = (\begin{smallmatrix}2\sqrt{2}&6\sqrt{2}+5\sqrt{3}\\\ \sqrt{3}&4\sqrt{2}+3\sqrt{3}\end{smallmatrix})$.

Quite generally, the matrix $M = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ satisfies $MA = BM$ if and only if $b=3a+5c$ and $d = 2a+3c$, and then $\det M = 2a^2 - 5c^2$. We can't solve $2a^2 - 5c^2 = \pm 1$ in ${\mathbf Z}$ (look at it mod 5), but we can solve it in ${\mathbf Z}[\sqrt{2}]$ using $a = \sqrt{2}$ and $c = 1$. That is how I found $U$. We can solve $2a^2 - 5c^2 = 1$ using $a = 2\sqrt{2}$ and $c = \sqrt{3}$, which is how I found $V$.

`$A=(\begin{smallmatrix}0&4\\\ 2&0\end{smallmatrix})$`

and`$B=(\begin{smallmatrix}0&8\\\ 1&0\end{smallmatrix})$`

are conj. by`$(\begin{smallmatrix}1&0\\\ 0&1/2\end{smallmatrix})$`

, but if they are conj. by`$(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix})$`

then $a=2d$ and $b=4c$, so $ad-bc=2d^2 - 4c^2$, which (contd.) $\endgroup$ – KConrad Apr 30 '12 at 2:56