This is a follow up question of this topic. Most of the following recap stems from the valuable answer therein.
Let $G,\overline G$ be groups, and $f\colon G\to\overline G$ a bijection. With reference to the following diagram: $$\begin{array}{c} G & \stackrel{f}{\longrightarrow} & \overline G \\ \downarrow{\mathfrak c} & & \downarrow{\mathfrak{\overline c}} \\ \operatorname{Sym}(G) & \stackrel{\psi_f}{\longleftarrow} & \operatorname{Sym}(\overline G) \end{array} $$ where $\mathfrak c$ and $\overline{\mathfrak c}$ are Cayley's embeddings and $\psi_f\colon\operatorname{Sym}(\overline G)\to\operatorname{Sym}(G)$ is defined by $\sigma\mapsto f^{-1}\sigma f$, the following results hold true: \begin{alignat}{1} &1)\space\forall g,h\in G: &&f(gh) = f(g)f(h) &&&\iff \mathfrak c=\psi_f\bar{\mathfrak c}f \\ &2)\space\exists\bar k\in\overline G,\forall g,h\in G: \space&&f(gh) = f(g)\bar kf(h) &&&\iff \operatorname{im}(\mathfrak c)=\operatorname{im}(\psi_f\bar{\mathfrak c}f) \\ \end{alignat} Both $1)$ and $2)$ define an equivalence relation in the set of all the groups of a given cardinal; in fact, the inverse $f^{-1}$ is still a map fulfilling $1)$ or $2)$. The equivalence classes are made of pariwise isomorphic groups, in case $1)$, and pairwise "quasi-isomorphic" (my definition) groups, in case $2)$. While as usual $G\cong\overline G$ denotes an isomorphism, let $G\approx\overline G$ denote a quasi-isomorphism. Since $2)$ boils down to $1)$ for the special case $\bar k=1_\overline G$, isomorphic groups are also quasi-isomorphic, but not necessarily viceversa: there might well be non-isomorphic groups which nonetheless turn out to be quasi-isomorphic. So, a quasi-isomorphism looks like an "isomorphism in a broader sense", though not to the point of equalling ("$\approx$") an abelian with a nonabelian group; in fact: \begin{alignat}{1} &(G\text{ abelian)}\wedge(f\text{ quasi-isomorphism)} &&\iff \\ &\exists\bar k\in\overline G,\forall g,h \in G: f(g)\bar kf(h)=f(h)\bar kf(g) &&\iff \\ &\exists\bar k\in\overline G,\forall \bar g,\bar h \in \overline G: \bar g\bar k\bar h=\bar h\bar k\bar g &&\iff \\ &\overline G\text{ is abelian} \\ \end{alignat} (for the last "$\iff$", see e.g. here).
As a first (minimal) test, does $C_4\stackrel{f}{\approx} K_4$? The answer is no, because, once set $C_4=\langle x\rangle$ and $f\colon C_4\to K_4$ the candidate quasi-isomorphism, we get: $f(x^2)=$ $\bar kf(x)^2=$ $\bar k$, whence $f(x^3)=$ $\bar kf(x^2)f(x)=$ $\bar k^2f(x)=$ $f(x)$, a contradiction because $x^3\ne x$.
Question. I'm looking for a pair of (possibly low order) non-isomorphic groups (necessarily both abelian or both nonabelian), which are quasi-isomorphic.
Edit. with reference to the following diagram: $$\begin{array}{c} \overline G & \stackrel{f'}{\longleftarrow} & G & \stackrel{f}{\longrightarrow} & \overline G \\ \downarrow{\mathfrak{\overline c}} & & \downarrow{\mathfrak c} & & \downarrow{\mathfrak{\overline c}} \\ \operatorname{Sym}(\overline G) & \stackrel{\psi_{f'}}{\longrightarrow} & \operatorname{Sym}(G) & \stackrel{\psi_f}{\longleftarrow} & \operatorname{Sym}(\overline G) \end{array} $$ the result in the answer can be written as: $$\operatorname{im}(\mathfrak c)=\operatorname{im}(\psi_f\bar{\mathfrak c}f)\Longrightarrow \mathfrak c=\psi_{f'}\bar{\mathfrak c}f'$$ where $f'(g):=f(1_G)^{-1}f(g)$.
Edit #2. For $\overline G=$ $(\mathbb R,+)$, such "quasi-isomorphism" seems instead to match the definition of quasi-morphism mentioned in the answer, with defect equal to $|f(1_G)|$.
