If $d = a_{i+1} - a_i$ is the constant difference of the given arithmetic sequence, we have $a_i = a_0 + d i$ for all $i \ge 0$. Because we are given $a_i > 0$ for all $i$, it follows that $a_0 > 0$ and $d \ge 0$.

If $d = 0$, the sequence is constant and any equiangular hexagon with side lengths $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ has all sides of length $a_0$ and so is a regular hexagon. There is only one such hexagon (up to congruence).

If $d > 0$, the sequence is not constant and the hexagons we consider have six sides of different lengths. Consider an equiangular hexagon with side lengths $a, b, c, a', b', c'$ (in this order) and extend the sides $b, a'$ and $c'$ so that we obtain a triangle as shown in the diagram.

Because all internal angles in an equiangular hexagon have a measure of $120^\circ$, the large triangle and the three small triangles are equilateral. Hence the side length of the large equilateral triangle is $b' + c' + a = a + b + c = c + a' + b'$. Subtracting $a$ from the first and $b$ from the second equation we see that the six side lengths of any equiangular hexagon satisfy

$$a + b = a' + b' \quad \text{and} \quad b + c = b' + c'. \qquad (1)$$

Subtracting these and rearranging gives us $a' + c = a + c'$ as well. We now show that for any six positive numbers $a, b, c, a', b', c'$ that satisfy (1), there exists an equiangular hexagon with these side lengths.

We essentially retrace the steps of our argument above: Equations (1) imply $b' + c' + a = a + b + c = c + a' + b'$. We then draw an equilateral triangle with this side length and cut off small equilateral triangles as shown in the diagram above. The resulting equiangular hexagon has the required side lengths.

Given $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$, the question is in how many ways can we order these numbers such that they satisfy (1). First note that the six numbers $a, b, c, a', b', c'$ satisfy (1) iff $a+K, b+K, c+K, a'+K, b'+K, c'+K$ satisfy (1). Therefore, it is sufficient to count the number of incongruent equiangular hexagons with side lengths $a_0, a_1, a_2, a_3, a_4, a_5$ in some order. For the same reason, we may assume $a_0 = 0$ so that $a_i = d i$. Moreover, the six numbers $a, b, c, a', b', c'$ satisfy (1) iff $d a, d b, d c, d a', d b', d c'$ do so. Hence, it is sufficient to study the case $d = 1$, i.e. $a_i = i$. In other words, we need to count the number of ways the numbers $0, 1, 2, 3, 4, 5$ can be ordered such that equations (1) hold. Cyclic changes of the order lead to congruent hexagons, hence we may assume $b = 5$.

Adding the equations (1) gives us $(a + b + c) + b = (a' + b' + c') + b'$, which we rewrite as $b - b' = (a' + b' + c') - (a + b + c) = 15 - 2(a + b + c)$. The last equality comes from $a + b + c + a' + b' + c' = 0 + 1 + 2 + 3 + 4 + 5 = 15$. Since we have chosen $b = 5$ to be the largest available number, we have $b - b' > 0$, hence $15 > 2(a + b + c)$, i.e. $a + b + c \le 7$ which means that $a + c \le 2$. Therefore, $\{a, c\} = \{0, 1\}$ and $\{a, c\} = \{0, 2\}$ are the only possibilities.

Swapping $a$ and $c$, as well as $a'$ and $c'$ leads to a congruent (via a reflection) equiangular hexagon. Therefore, the two possibilities (up to congruence) are $(a, c) = (0, 1)$ and $(a, c) = (0, 2)$.

When $(a, c) = (0, 1)$ and $b = 5$, we have $a + b + c = 6$ and so $a' + b' + c' = 15 - 6 = 9$. Therefore, $b' = (a + b + c) + b - (a' + b' + c') = 6 + 5 - 9 = 2$. Hence, $a' = a + b - b' = 0 + 5 - 2 = 3$ and $c' = c + b - b' = 1 + 5 - 2 = 4$. We obtain $(a, b, c, a', b', c') = (0, 5, 1, 3, 2, 4)$.

Similarly, starting with $(a, c) = (0, 2)$ we obtain $a+b+c=7$ and $a'+b'+c'=8$. This leads to $b'=4$, $a'=1$ and $c'=3$ and $(a, b, c, a', b', c') = (0, 5, 2, 1, 4, 3)$.

Since $a + b + c = 0 + 5 + 1 = 6$ from the first solution is neither equal to $a + b + c = 0 + 5 + 2 = 7$ nor to $b + c + a' = 5 + 2 + 1 = 8$ from the second solution, the two solutions do not lead to congruent hexagons.

Translating this back to the original set-up, we have shown that (in case $d > 0$) for each $i \ge 0$ the six numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ can be ordered in exactly two ways (up to congruence of the hexagons) such that these numbers are the side lengths of an equiangular hexagon in this order, namely

$$a_i, a_{i+5}, a_{i+1}, a_{i+3}, a_{i+2}, a_{i+4} \quad \text{and} \quad a_i, a_{i+5}, a_{i+2}, a_{i+1}, a_{i+4}, a_{i+3}.$$