Fibonacci sequence ($\mathbb{F}$) has a repeating cycle known as Pisano number $\pi\text{(x)}$ , when $mod \text{ x}$ is applied upon the sequence. Length of the cycles can be found from: http://oeis.org/A001175
With $\text{mod 9}$ cycle length $\pi\text{(9)}$ is $24$, so the repeating 24-tuple would be:
$$\mathbb{S} = \mathbb{F}_{n < 25} \text{ mod }9 = (0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1)$$
There are several doubling sub sequences inside $\mathbb{S}$. For example:
$$\mathbb{S}_{4n-5} = (1,2,4,8,7,5)$$
and reverse of it:
$$\mathbb{S}_{4n+1} = (5,7,8,4,2,1)$$
On the other hand doubling multiset is:
$$\mathbb{D} = 2^n = [1,2,4,8,16,32,64,128,...]$$
Now my attention is on the fact that the doubling sequence $\text{mod 9}$ also happens to have a same repeating 6-tuple with $\mathbb{S}_{4n-5}$, namely:
$$\mathbb{D}_{n<6} \text{ mod 9}= (1,2,4,8,7,5) = \mathbb{S}_{4n-5}$$
So my question is if this forward and reverse doubling subsequence inside the Fibonacci $\text{mod 9}$ sequence is purely mathematical coincidence or if it can be explained by some moderate algebra?
References:
http://webspace.ship.edu/msrenault/fibonacci/fib.htm
Chart for visualization
Given $\varphi=\frac{1+\sqrt{5}}{2}$, the minimal polynomial of $\varphi^4$ over $\mathbb{Q}$ is given by $$ q(x) = x^2-7x+1 \equiv (x+1)^2 \pmod{9} $$ hence it follows that for any $k\in[1,4]$, the sequences defined by $G_{n}=F_{4n+k}$ behave like: $$ G(n) \equiv (-1)^n\cdot\left( A+Bn\right) \pmod{9}.$$ In particular, $$ F_{4n+1}\equiv (-1)^n (1+3n)\pmod{9},\qquad F_{4n-5}\equiv (-1)^n (-4+3n)\pmod{9} $$ that easily explains your reversing property.