Let $G$ be a group, and assume that there exist $a, b, c \in G$ such that $abc$, $acb$, $bac$, $bca$, $cab$ and $cba$ are precisely 5 distinct elements (i.e. that precisely two of the products are equal).

**Question 1:** Does it follow that there exist $d, e, f \in G$ such that
$def$, $dfe$, $edf$, $efd$, $fde$ and $fed$ are precisely 4 distinct elements?
And if not -- does the non-existence of such $d, e, f \in G$ at least imply
that $G$ is infinite?

*Remark:* When one replaces 5 by 6, the answer is *no*. -- The smallest group
which can be taken as an example here has order 54. It is
$$
G_{54,8} :=
\langle (1,4,7)(2,5,8)(3,6,9), (3,4,5)(6,8,7), (3,6)(4,7)(5,8) \rangle.
$$

**Question 2:** Let $G$ be as above, and assume further that there are no
$d, e, f \in G$ such that $def$, $dfe$, $edf$, $efd$, $fde$ and $fed$ are
pairwise distinct. If $G$ is finite, does it follow that the order of $G$
is a multiple of 5?

*Remark:* The groups of order up to 625 which fulfill the conditions have orders
20, 40, 60, 80, 100, 120, 125, 140, 160, 180, 200, 220, 240, 250, 260, 280,
300, 320, 340, 360, 375, 380, 400, 420, 440, 460, 480, 500, 520, 540, 560,
580, 600, 620 and 625, respectively.

*Side note:* A related
earlier question of mine
remains unsolved so far.

**Added on Aug 21, 2018:** Given a group $G$, put
$$
{\rm P}_3(G) :=
\left\{ |\{abc, acb, bac, bca, cab, cba\}| \ \big| \ a,b,c \in G \right\}.
$$
Then clearly we have ${\rm P}_3(G) = \{1\}$ if and only if $G$ is abelian.
Computational investigations further suggest that ${\rm P}_3(G)$ is always one
of $\{1\}$, $\{1,2\}$, $\{1,2,3\}$, $\{1,2,3,4\}$, $\{1,2,3,4,5\}$,
$\{1,2,3,4,5,6\}$, $\{1,2,3,4,6\}$, $\{1,2,3,6\}$ and $\{1,2,6\}$ --
where $\{1,2,3,4\}$, $\{1,2,3,4,5,6\}$ and $\{1,2,3,4,6\}$ are all very
common, while $\{1,2,3,6\}$ and $\{1,2,3,4,5\}$ impose more-or-less
severe restrictions on the structure of the group.

5more comments