Direct limits of a matrix and its transpose Let $A \in M_n(\mathbb Z)$ and $A^T$ denote the transpose of $A$. Define the direct limits
$$H_1 = \mathrm{colim} (\mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \mathbb Z^n \xrightarrow{A} \dots)$$
and
$$H_2 = \mathrm{colim} (\mathbb Z^n \xrightarrow{A^T} \mathbb Z^n \xrightarrow{A^T} \mathbb Z^n \xrightarrow{A^T} \dots).$$

QUESTION. Is it true that $H_1 \cong H_2$? If not, are there nice counterexamples?

Remarks. If $z = \det A = \det A^T$ then $H_1, H_2 \subseteq \mathbb Z[\frac{1}{z}]^n$. Therefore, a counterexample would need the have strict inclusions. Also, in the examples we have looked at, if the eigenvalues $\lambda_1, \ldots, \lambda_n \in \mathbb C$ of $A$ are integers then we find $H_1 \cong \mathbb Z[\frac{1}{\lambda_1}] \oplus \dots \oplus \mathbb Z[\frac{1}{\lambda_n}] \cong H_2$ (though we don't claim this to be true in general).
Further remarks.
Thank you to David Handelman for the answer and the example. The matrices we are interested in arise in the context of substitutional tilings. In most cases we were able to calculate the direct limits $H_1$ and $H_2$ using ad hoc methods. We did however find a matrix for which we were unable to calculate $H_1$: That matrix is $A = \left(
\begin{array}{ccccc}
 5 & -2 & 2 & -4 & -2 \\
 2 & 0 & 2 & -6 & -5 \\
 5 & -3 & 5 & -11 & -9 \\
 -2 & 1 & 1 & -3 & -3 \\
 3 & -1 & 0 & 0 & 0 \\
\end{array}
\right)$. We tried to guess that $H_1$ is isomorphic to $\mathbb Z^2 \oplus \mathbb Z[\frac{ 1 }{ 6 }]^3$ but we couldn't show this (it might be false). Unfortunately we're not well versed in algebraic number theory, so we are not sure how to decide whether $H_1$ is isomorphic to $H_2$ for this particular matrix.
 A: No. Define the matrix [nasty comment about LaTeX deleted]
$$
A =\pmatrix {3 & 7 \\ 2 & 1 \\}. 
$$
A left  eigenvector for the eigenvalue $c = 2 + \sqrt{15}$ is $(2,\sqrt{15}-3)$, whose entries generate the maximal ideal sitting over $2$; it is non-principal in $R = \mathbf{Z}[\sqrt {15}]$ (since $a^2- 15b^2 = \pm 2$ is unsolvable in $\mathbf{Z}$), and a little more is true. A right eigenvector is $(7, \sqrt{15}-3)^T$, whose entries generate a principal ideal. 
The basic idea is to use the fact that the eigenvalues are $c= 2+ \sqrt 15$ and $d =2- \sqrt{15}$, either one generates $R = \mathbf{Z}[\sqrt{15}]$, the ring of integers in $\mathbf{Q}[\sqrt 15]$; from the right eigenvector,  thus $\mathbf{Z}^2$ as a left $R$-module corresponds to a principal ideal; on the other hand, $\mathbf{Z}^2$ as right $R$-module is not principal. Moreover, it remains non-principal in $R[c^{-1}] = \mathbf{Z}[c^{-1}]$. Finally, the limiting groups are strongly indecomposable, so that isomorphism as abelian groups entails isomorphism as $R$-modules.  
Set $G$ to be the stationary direct limit determined by $A$ and $G’$ that determined by $A^T$. We need a few (interesting) results, which are sketched below. 
(a) $G$ contains no elements of infinite height. Follows easily from the fact that the content of each column in all powers of the matrix is  one. 
(b) $G$ is strongly indecomposable. If not, there would exist a subgroup $H = C \oplus D$ of finite index. Since the rank of $G$ is two, each of $C$ and $D$ is isomorphic to a subgroup of the rationals. Since neither can include elements of infinite height, each must be free. But then $G$ would be free, which it clearly isn’t (since the determinant of $A$ is nonzero but not $\pm1$). [There are more general sufficient  conditions for a stationary limit group to be strongly indecomposable; e.g., for size two, determinant not zero, one, or minus one, and the trace and determinant be relatively prime.]
(c) Let $E$ denote the endomorphism ring of $G$. Let $\alpha$ denote the automorphism of $G$ induced by $A$ (explcitly, $[v,m] \mapsto [Av,m]$; its inverse is $[v,m] \to [v,m+1]$). Then every endomorphism of $G$ commutes with $\alpha$. To see this, note that $\mathbf{Q}[A]$ is a two-dimensional field inside $M_2 \mathbf{Q}$ (isomorphic to $\mathbf{Q}[\sqrt{15}]$), hence is a maximal $\mathbf{Q}$-subalgebra thereof. Thus if $e \in E$ does not commute with $\alpha$, there exists an element $f$ of $E \otimes \mathbf{Q}$ that is idempotent. Hence there exists an integer $N$ such that $Nf \in E$, and then the range of $Nf$ is rank one, and so is that of $N(1-f)$. This yields a direct sum of finite index in $G$, contradicting strong indecomposability.
(d) The centralizer in $M_2 \mathbf{Z}$ of $A$ includes $\mathbf{Z}[A]$, and since this is isomorphic to an integrally closed ring, $\mathbf{Z}[\sqrt {15}]$, we must have that the centralizer is $\mathbf{Z}[A]$. Then it is easy to check that  $E = \mathbf{Z}[c^{-1}]$.
(e) We also see that if $d = 2 - \sqrt {15}$, then $d^{-1} \not\in E$ (elementary). 
(f) Any (group) isomorphism between the strongly indecomposable groups $G$ and $G’$ induces an isomorphism of their endomorphism rings. Hence if $G$ and $G’$ were isomorphic, then they would be isomorphic up to an automorphism of $E$. But the only candidate for an automorphism is the Galois automorphism, which is not defined on $E$, because $d^{-1} \not\in E$. Hence any group isomorphism would be an $E$-module isomorphism.
(g) Since the right eigenvector corresponds to a principal ideal, $G$ is isomorphic the principal ideal, $E$ itself. On the other hand, the left eigenvector of $A$ yields the ideal $I= (2,\sqrt{15}-3)$ of $R$, and this remains non-principal in $E$. Thus $G’$ is isomorphic to  $I\cdot E$, which is not a free $E$-module, so not isomorphic to $G$ as $E$-modules, and therefore by (f), is not isomorphic to $G$ as abelian groups.
Historical comment In the context of partially ordered direct limits (where the matrix $A$ is primitive, that is, all the entries are nonnegative, and in some power, they are all strictly positive), this type of construction has appeared in connection with classification up to shift equivalence of shifts of finite type (dynamical systems appearing as subshifts of full shifts, in topological dynamics) since the 70s. Here, the invariant includes the induced action of $A$, and it was well-known that there are size two matrices $A$ for which $A$ is not shift equivalent to its transpose. We cannot just adapt these examples directly, since the order data (the positivity) is crucial; for example, we could have in this context $|det\ A|= 1$, so that the underlying group of the limit is free.  
Instead, we use strong indecomposability to restrict considerably the endomorphisms (in the ordered case, the natural set of endomorphisms are those that are continuous).
