Yes. Let $H$ be the commutative group $H=\mathbb{F}_{2}^{3}$, where $\mathbb{F}_{2} \cong \mathbb{Z} / 2 \mathbb{Z}$ is the field with two elements. The group of automorphisms of $H$ is the general linear group $\mathrm{GL}_{3} \mathbb{F}_{2}$; it has $$(8-1) \cdot(8-2) \cdot(8-4)=7 \cdot 6 \cdot 4=168$$ elements. One of them is the shift operator $\phi:\left(x_{1}, x_{2}, x_{3}\right) \mapsto\left(x_{2}, x_{3}, x_{1}\right)$. Now let $T=\left\{a^{0}, a^{1}, a^{2}\right\}$ be a group of order 3 (written multiplicatively); it acts on $H$ by $\tau(a)=\phi$. Let $G$ be the semidirect product $G=H \rtimes_{\tau} T$. In other words, $G$ is the group of 24 elements $$G=\left\{b a^{i}: \quad b \in H, i \in(\mathbb{Z} / 3 \mathbb{Z})\right\}, \quad a b=\phi(b) a$$ $G$ has one element $e$ of order 1 and seven elements $b, b \in H, b \neq e$ of order 2 If $g=b a$, we find that $g^{2}=b a b a=b \phi(b) a^{2} \neq e$, and that $$g^{3}=b \phi(b) a^{2} b a=b \phi(b) a \phi(b) a^{2}=b \phi(b) \phi^{2}(b) a^{3}=\psi(b)$$ where the homomorphism $\psi: H \rightarrow H$ is defined as $\psi:\left(x_{1}, x_{2}, x_{3}\right) \mapsto\left(x_{1}+x_{2}+x_{3}\right)(1,1,1)$. It is clear that $g^{3}=\psi(b)=e$ for 4 elements $b \in H$, while $g^{6}=\psi^{2}(b)=e$ for all $b \in H$. We see that $G$ has 8 elements of order 3, namely $b a$ and $b a^{2}$ with $b \in \operatorname{Ker} \psi$, and 8 elements of order 6 namely $b a$ and $b a^{2}$ with $b \notin \operatorname{Ker} \psi$. That accounts for orders of all elements of $G$. Let $b_{0} \in H \backslash \operatorname{Ker} \psi$ be arbitrary; it is easy to see that $G$ is generated by $b_{0}$ and $a$. As every automorphism of $G$ is fully determined by its action on $b_{0}$ and $a$, it follows that $G$ has no more than $$7 \cdot 8=56$$ automorphisms. Remark. $G$ and $H$ can be equivalently presented as subgroups of $S_{6}$, namely as $H=\langle(12),(34),(56)\rangle$ and $G=\langle(135)(246),(12)\rangle$