Let $G$ be a finite group of order $16$ acting on a finite set $S$ of size $5$. Suppose we know that there are $< 4$ distinct $G$-orbits. We now work out step-by-step the possible values for the NUMBER of orbits of this $G$-action.

We begin with a general observation. Let $X$ be any non-empty set. Since the size of $G$ is the $4$-th power of the prime $2$, by the orbit-stabilizer theorem the possible values for orbit sizes of a $G$-action on $X$ are



Next, we specialize to the case where $G$ acts on our given set $S$ of size $5$. We can his further limits the possible values for the orbit size to



And since distinct orbits are pairwise disjoint and their union is the set $S$, if we denote by

$A =$number of size $2^0$ orbits
$B =$number of size $2^1$ orbits
$C =$number of size $2^2$ orbits

then

$(\ast 1)$ $5 = A \times 1 + B \times 2 + C \times 4$

Don't forget the initial hypothesis

$(\ast 2)$ there are $< 4$ distinct $G$-orbits.

We are interested in the total number of distinct $G$-orbits, which is equal to $A + B + C$. Since $5 < 2\times 2^2$, it follows from $(\ast1)$ that

$(\ast 3)$ $0 \le A \le 5$, $0 \leq B \leq 2$, and $0 \leq C \leq 1$.

To finish the problem we can now proceed by brute force, taking each possible triple of $A, B, C$ given by $(\ast 3)$ and see if it satisfies $(\ast 2)$. Here is a more systematic approach.

First, if $C=1$ then $(\ast)$ becomes

$1 = A\times 1 + B\times 5.$

We are still subject to condition $(\ast 3)$, so

If $C=1$ then $A$ = and $B$ =

For the rest of this argument we take $C=0$, in which case $(\ast 1)$ and $(\ast 2)$ become

$5 = A + B \times 2$.
$10 > A + B.$

Consequently,

If $C=0$ then $A$ = and $B$ =