Shuffling the digits of an integer so that the ratio between the resulting numbers is fixed.
We want to find triples of integers $(x,y,z)$ such that $$0\lt x\le 9, 0\le y\le 9,0\le z\le 9\tag1$$ $$x\not=y,y\not=z,z\not=x\tag2$$ $$\frac{100z+10x+y}{100y+10z+x}=\frac{100y+10z+x}{100x+10y+z}\tag3$$
From $(3)$, we get
$$ yz-x^2=10(xz-y^2)\implies yz\equiv x^2\pmod{10}\tag4$$ and $$x=-5z+\sqrt{25z^2+10y^2+yz}\implies \sqrt{25z^2+10y^2+yz}\ \in\mathbb Z\tag5$$
Also, if $y\lt z\lt x$, then $yz-x^2=10(xz-y^2)$ does not hold since LHS is negative while RHS is positve.
If $x\lt z\lt y$, then $yz-x^2=10(xz-y^2)$ does not hold since LHS is positive while RHS is negative.
If $x^2\equiv 1\pmod{10}$, then $yz=21$ and so $(y,z)=(3,7),(7,3)$ each of which doesn't satisfy $(5)$.
If $x^2\equiv 4\pmod{10}$, then $yz=4,14,24,54$ and so $(y,z)=(1,4),(4,1),(7,2),(3,8),(4,6),(6,4),(6,9),(9,6)$ where only $(y,z)=(6,4),(9,6)$ satisfy $(5)$ with $x=8,12$ respectively. (we don't need to consider the case $(y,z)=(2,7)$ since then $x=8$ for which $y\lt z\lt x$ holds. Also, we don't need to consider the case $(y,z)=(8,3)$ since then $x=2$ for which $x\lt z\lt y$ holds.)
If $x^2\equiv 9\pmod{10}$, then $yz=9$ and so $(y,z)=(1,9),(9,1)$ each of which doesn't satisfy $(5)$.
If $x^2\equiv 6\pmod{10}$, then $yz=6,16,36,56$ and so $(y,z)=(1,6),(6,1),(3,2),(2,8),(8,2),(4,9),(9,4),(7,8)$ where only $(y,z)=(3,2)$ satisfies $(5)$ with $x=4$. (we don't need to consider the case $(y,z)=(2,3)$ since then $x=4$ or $6$ for which $y\lt z\lt x$ holds. Also, we don't need to consider the case $(y,z)=(8,7)$ since then $x=4$ or $6$ for which $x\lt z\lt y$ holds.)
If $x^2\equiv 5\pmod{10}$, then $x=5$ and one of $y,z$ is $5$, which does not satisfy $(2)$.
So, the only solutions are $$\color{red}{(x,y,z)=(4,3,2),(8,6,4)}$$ where we have $$\frac{243}{324}=\frac{324}{432}=\frac 34=\frac{486}{648}=\frac{648}{864}$$
Not a complete answer but can be open for further discussion
See one famous example for cyclic number:
$$\frac{1}{7}=0.\overline{142857}$$
we have
$$142857:285714:428571:571428:714285,857142=1:2:3:4:5:6$$
You may pick up $1:2:4$ for a case of $6$ digits.
In general for base $b$, the cyclic-permutation decimal can be in the form of
$$\frac{b^{p-1}-1}{p}$$
Besides the cyclic numbers mentioned in the above link, a number of subsets of $\dfrac{k}{37}$ give further kinds of cyclic numbers (including $0$):
$k\in \{ 1,10,26 \}$ gives $0.\overline{027},0.\overline{270},0.\overline{702}$
$k\in \{ 2,15,20 \}$ gives $0.\overline{054},0.\overline{405},0.\overline{540}$
$k\in \{ 3,4,30 \}$ gives $0.\overline{081},0.\overline{108},0.\overline{810}$
$k\in \{ 5,13,19 \}$ gives $0.\overline{135},0.\overline{351},0.\overline{513}$
$k\in \{ 6,8,23 \}$ gives $0.\overline{162},0.\overline{216},0.\overline{621}$
$k\in \{ 7,33,34 \}$ gives $0.\overline{162},0.\overline{216},0.\overline{621}$
$k\in \{ 9,12,16 \}$ gives $0.\overline{243},0.\overline{324},0.\overline{432}$
$k\in \{ 11,27,36 \}$ gives $0.\overline{297},0.\overline{729},0.\overline{972}$
$k\in \{ 14,29,31 \}$ gives $0.\overline{378},0.\overline{783},0.\overline{837}$
$k\in \{ 17,22,35 \}$ gives $0.\overline{459},0.\overline{594},0.\overline{945}$
$k\in \{ 18,24,32 \}$ gives $0.\overline{486},0.\overline{648},0.\overline{864}$
$k\in \{ 21,25,28 \}$ gives $0.\overline{567},0.\overline{675},0.\overline{756}$
Only subsets $\{ 9,12,16 \}$ and $\{ 18,24,32 \} $ are your favourable cases.
Note that $27\times 37=999$
For $k=3n$, $$\frac{k}{27} = 0.\overline{n}$$
For other $k$ values, there're $6$ triplets of cyclic numbers but there're no favourable cases here.