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The solvable Baumslag-Solitar group $BS(1, n) = \langle a, b \, \vert \, aba^{-1} = b^n \rangle$ with $n \in \mathbb{Z} \setminus \{0\}$, has only one $T$-system of generating pairs and any number of Nielsen classes, including infinity, can be achieved for a suitable choice of $n$. This [preprint][9] presents of proof of my claim, see Corollary E.

[1] "Über die gruppen $A^{a}B^{b}=1$", O. Schreier, 1923.
[2] "Über die Nielsensche Kürzungsmethode in freien Produkten mit Amalgam", H. Zieschang, 1970.
[3] "Presentations of the trefoil group", M. J. Dunwwoody, A. Pietrowsky, 1973.
[4] "Transitivity systems of some one-relator groups", A. M. Brunner, 1974.
[5] "Generators of the free product with amalgamation of two infinite cyclic groups", H. Zieschang, 1977.
[6] "Automorphisms and Hopficity of certain Baumslag-Solitar groups", D. J. Collins and F. Levin, 1983.
[7] "Tree actions of automorphism groups", N. D. Gilbert et al., 2000.
[8] "Nielsen equivalence in closed surface groups", L. Louder, 2010, [arXiv:1009.0454][9].
[9]: http://arxiv.org/abs/1604.08896

The solvable Baumslag-Solitar group $BS(1, n) = \langle a, b \, \vert \, aba^{-1} = b^n \rangle$ with $n \in \mathbb{Z} \setminus \{0\}$, has only one $T$-system of generating pairs and any number of Nielsen classes, including infinity, can be achieved for a suitable choice of $n$. This preprint presents of proof of my claim, see Corollary E.

[1] "Über die gruppen $A^{a}B^{b}=1$", O. Schreier, 1923.
[2] "Über die Nielsensche Kürzungsmethode in freien Produkten mit Amalgam", H. Zieschang, 1970.
[3] "Presentations of the trefoil group", M. J. Dunwwoody, A. Pietrowsky, 1973.
[4] "Transitivity systems of some one-relator groups", A. M. Brunner, 1974.
[5] "Generators of the free product with amalgamation of two infinite cyclic groups", H. Zieschang, 1977.
[6] "Automorphisms and Hopficity of certain Baumslag-Solitar groups", D. J. Collins and F. Levin, 1983.
[7] "Tree actions of automorphism groups", N. D. Gilbert et al., 2000.
[8] "Nielsen equivalence in closed surface groups", L. Louder, 2010, arXiv:1009.0454.

The proof is elementary and can be summarized as follows. First identify $G = BS(1, n)$ with $\mathbb{Z}[\frac{1}{n}] \rtimes_n \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts as the multiplication by $n$ on $R = \mathbb{Z}[\frac{1}{n}]$. Then the commutator of a generating pair of $G$ is of the form $(1 - n)u$ for some unit $u$ of $R$ and every unit arises in this way. By Higman's lemma, if two generating pairs $\mathbf{g}_1$ and $\mathbf{g}_2$ of $G$ are Nielsen equivalent then the corresponding units $u_1$ and $u_2$ are related by an element of the form $(-1)^kn^l$ with $k, l \in \mathbb{Z}$. It is not difficult to show that every generating pair of $G$ can be Nielsen reduced to $(a, b')$ for some $b'$ that identifies with a unit of $R$. Now it is immediate to see that if $u_1$ and $u_2$ are related by an element of $(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$ then $\mathbf{g}_1$ and $\mathbf{g}_2$ are Nielsen equivalent. So $G$ has as many Nielsen classes of generating pairs as elements in $R^{\times}/(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$. A careful inspection of $R^{\times}$ shows that the cardinality of the latter is infinite if $n = 6$ and equal to $d \in \mathbb{N} \setminus \{0\}$ if $n = 2^d$. As the group $G$ has only one $T$-system of generating pairs (see e.g, [4]), it answers OP's question with residually finite instances.

The proof is elementary and can be summarized as follows. First identify $G = BS(1, n)$ with $\mathbb{Z}[\frac{1}{n}] \rtimes_n \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts as the multiplication by $n$ on $R = \mathbb{Z}[\frac{1}{n}]$. Then the commutator of a generating pair of $G$ is of the form $(1 - n)u$ for some unit $u$ of $R$ and every unit arises in this way. By Higman's lemma, if two generating pairs $\mathbf{g}_1$ and $\mathbf{g}_2$ of $G$ are Nielsen equivalent then the corresponding units $u_1$ and $u_2$ are related by an element of the form $(-1)^kn^l$ with $k, l \in \mathbb{Z}$. It is not difficult to show that every generating pair of $G$ can be Nielsen reduced to $(a, b')$ for some $b'$ that identifies with a unit of $R$. Now it is immediate to see that if $u_1$ and $u_2$ are related by an element of $(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$ then $\mathbf{g}_1$ and $\mathbf{g}_2$ are Nielsen equivalent. So $G$ has as many Nielsen classes of generating pairs as elements in $R^{\times}/(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$. A careful inspection of $R^{\times}$ shows that the cardinality of the latter is infinite if $n = 6$ and equal to $d \in \mathbb{N} \setminus \{0\}$ if $n = 2^d$. As the group $G$ has only one $T$-system of generating pairs by [Theorem 2.4, 4], it answers OP's question with residually finite instances.

A. M. Brunner proved in [4] that for each $n \ge 0$ the generating pair $(a, b^{2^n})$ of the non-Hopfian Baumslag-Solitar group $BS(2, 3) = \langle a, b \, \vert \, ab^2a^{-1} = b^3 \rangle$ lies in a different $T$-system of $BS(2, 3)$ but he left open the question whether each $T$-system has a representative of this form. By [Theorem B, 7], every automorphism of $BS(2, 3)$ is inner except $(a, b) \mapsto (a, b^{-1})$. Therefore each $T$-system contains at most two Nielsen equivalence classes and the $T$-system of $(a, b^{2^n})$ contains a single Nielsen equivalence class. More generally, if $p, q > 1$ and neither divides the other, then $\Gamma = \operatorname{Out}(BS(p, q))$ is isomorphic to the dihedral group of order $d = 2\vert p - q \vert$ so that every $T$-system of generating pairs contains at most $d$ Nielsen equivalence classes and the $T$-system of $(a, b)$ contains a single Nielsen equivalence class. If $p$ properly divides $q$, then $\Gamma$ is not finitely generated [6]. But also in this case, inspection the automorphism group shows that the $T$-system of $(a, b)$ contains a single Nielsen equivalence class.

A. M. Brunner proved in [4] that for each $n \ge 0$ the generating pair $(a, b^{2^n})$ of the non-Hopfian Baumslag-Solitar group $BS(2, 3) = \langle a, b \, \vert \, ab^2a^{-1} = b^3 \rangle$ lies in a different $T$-system of $BS(2, 3)$ but he left open the question whether each $T$-system has a representative of this form. By [Theorem B, 7], every automorphism of $BS(2, 3)$ is inner except the automorphism induced by $(a, b) \mapsto (a, b^{-1})$. Therefore each $T$-system contains at most two Nielsen equivalence classes and the $T$-system of $(a, b^{2^n})$ contains a single Nielsen equivalence class. More generally, if $p, q > 1$ and neither divides the other, then $\Gamma = \operatorname{Out}(BS(p, q))$ is isomorphic to the dihedral group of order $d = 2\vert p - q \vert$ generated by the images of $(a, b) \mapsto (ab, b)$ and $(a, b) \mapsto (a, b^{-1})$. As a result every $T$-system of generating pairs contains at most $d$ Nielsen equivalence classes and the $T$-system of $(a, b)$ contains a single Nielsen equivalence class. If $p$ properly divides $q$, then $\Gamma$ is not finitely generated [6]. But also in this case, inspection the automorphism group shows that the $T$-system of $(a, b)$ contains a single Nielsen equivalence class.