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The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has one only Nielsen equivalence class of generating pairs for $n \in \{2, 3, 4, 6\}$. The group $\text{Heisenberg}_2(\mathbf{Z}/2\mathbf{Z})$ is the dihedral group of order $8$, which is not a quotient of $F_2$ by a characteristic subgroup. But $B(2, 3) = \text{Heisenberg}_2(\mathbf{Z}/3\mathbf{Z})$, is certainly another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has one only Nielsen equivalence class of generating pairs for $n \in \{2, 3, 4, 6\}$. The group $\text{Heisenberg}_2(\mathbf{Z}/2\mathbf{Z})$ is the dihedral group of order $8$, which is not a quotient of $F_2$ by a characteristic subgroup. But $B(2, 3) = \text{Heisenberg}_2(\mathbf{Z}/3\mathbf{Z})$, is certainly another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has only Nielsen equivalence class of generating pairs for $n \in \{2, 3\}$. For $n = 2$, we obtain the wreath product $C_2 \wr C_2$, the other non-abelian group with $8$ elements answering the question. For $n = 3$, we get $B(2, 3)$, another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has one only Nielsen equivalence class of generating pairs for $n \in \{2, 3, 4, 6\}$. The group $\text{Heisenberg}_2(\mathbf{Z}/2\mathbf{Z})$ is the dihedral group of order $8$, which is not a quotient of $F_2$ by a characteristic subgroup. But $B(2, 3) = \text{Heisenberg}_2(\mathbf{Z}/3\mathbf{Z})$, is certainly another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has only Nielsen equivalence class of generating pairs for $n \in \{2, 3\}$. For $n = 2$, we obtain the wreath product $C_2 \wr C_2$, which the other non-abelian group with $8$ elements which answers the question. For $n = 3$, this is $B(2, 3)$, which is then another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has only Nielsen equivalence class of generating pairs for $n \in \{2, 3\}$. For $n = 2$, we obtain the wreath product $C_2 \wr C_2$, the other non-abelian group with $8$ elements answering the question. For $n = 3$, we get $B(2, 3)$, another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.

[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.