You want to count each collection of hexagons interrelated by rotations as a single "truly different" hexagon. If $X$ is the set of hexagons, and $G$ is the group of rotations, then you want to find $|X/G|$: the number of orbits of elements of $X$ under $G$. Burnside's lemma gives the equation for this:
$$
|X/G| = \frac{1}{|G|}\sum_{g \in G}|X^{g}| = \frac{|X|}{|G|} + \frac{1}{|G|}\sum_{g \in G | g \neq e}|X^g|,
$$
where $X^g$ is the set of elements fixed by $g$, and the second sum excludes the identity element of $G$. In this case, you correctly counted $|G|=6$ and $|X|=90$, and so if no hexagons were fixed by a non-zero rotation, then $90/6=15$ would be the right answer. However, there are $6$ hexagons fixed by the $180^{\circ}$ rotation, giving an additional contribution of $6/6=1$, for a total of $15+1=16$.
$A$, $B$, $C$, $D$ and $E$ are five persons who are to be seated around a circular table such that $A$ and $B$ must sit together and $C$ and $D$ must never sit together. In how many ways can they be seated?
There are four possible seating arrangements.
Seat E. Since A and B sit together and C and D are separated, C and D must both be adjacent to E. Therefore, choosing whether C or D sits to E's immediate left also determines who sits to E's immediate right and choosing whether A or B sits two seats to E's left also determines who sits two seats to E's right. Hence, there are $2 \cdot 2 = 4$ permissible seating arrangements, as shown below.
Notice that none of these seating arrangements can be obtained from another by rotation.
Should the rotation of a particular arrangement be construed as the same or different?
By convention, a rotation of a particular arrangement is considered to be the same unless the seats are labeled or we are given a particular reference point (such as a special chair or the north end of the table).
Notice that we have already accounted for rotational invariance by measuring our seating arrangements relative to the position of E.
Say we have $10$ persons sitting around a circular table, with $3$ of them wanting to sit together and $4$ other persons who wish to be separated? In how many ways can they be seated?
We use the block of three people who wish to sit together as our reference point. Say the people are $A$, $B$, and $C$. In how many ways can they be arranged within the block?
$3!$
Suppose the four people who wish to be separated are $D$, $E$, $F$, and $G$. Since there are only seven seats left at the table, they must be seated in the seats that are $1$, $3$, $5$, and $7$ positions to the left of the block. In how many ways can they be seated?
$4!$
Let's call the remaining three people $G$, $H$, and $I$. In how many ways, can they be seated in the remaining three chairs?
$3!$
Therefore, by the Multiplication Principle, the number of permissible seating arrangements is
$3!4!3!$
Best Answer
Converting my comment to an answer to get this off the Unanswered list:
Your second version is correct: there are $6$ gaps, not $5$, so there are $6(6-1)!=6\cdot5!=6!$ ways to insert the $7$-th object into the circle of $6$ objects. This is of course the same as the result of the original $(7-1)!$ calculation.