Its not a coincidence. It is a general result for any number represented in a positive base $b$, its representation in base $b^k$ for some positive integer $k$, is simply grouping $k$ digits in its base $b$ representation starting from the least significant digit to the most significant digit.
Take base $10$ for example.
$427428$ in base $10$ is $(4,2,7,4,2,8)$.
$427428$ in base $10^2$ is $(42,74,28)$.
$427428$ in base $10^3$ is $(427,428)$.
You can now find an analogy to your problem.
Edit: Proof for the above claim
Let a number $n$ be represented in base $b$ positional system as
$n = (a_{m-1},a_{m-2},\space...\space,a_1,a_0)$ such that $0 \le a_i \lt b \space \forall i\in{0,1,\space...\space,m-1}$
where $m$ represents the number of digits $n$ has in the base $b$ representation. We wish to represent $n$ in base $B = b^k$ for some positive integer $k$.
For the sake of simplicity of the argument, let $m$ be a multiple of $k$, i.e. $m = c k$ for some integer $c$. (If $m$ is not a multiple of $k$, the representation of $n$ in base $b$ can be prefixed by $0$'s untill the number of digits are a multiple of $k$)
From the representation in base $b$, we can write $n$ as follows
$n = a_{m-1} b^{m-1} + a_{m-2} b^{m-2} + \space ... \space + a_2 b^2 + a_1 b + a_0$
$\space\space= a_{ck-1} b^{ck-1} + a_{ck-2} b^{ck-2} + \space ... 945
\space + a_2 b^2 + a_1 b + a_0$
$\space\space = a_{ck-1}b^{ck-1} + a_{ck-2}b^{ck-2} + \space ... \space + a_{(c-1)k+1} b^{(c-1)k+1} + a_{(c-1)k} b^{(c-1)k}$
$\space\space\space + a_{(c-1)k-1}b^{(c-1)k-1} + a_{(c-1)k-2}b^{(c-1)k-2} + \space ... \space + a_{(c-2)k+1} b^{(c-2)k+1} + a_{(c-2)k} b^{(c-2)k}$
$\space\space\space\vdots$
$\space\space\space + a_{2k-1}b^{2k-1} + a_{2k-2}b^{2k-2} + \space ... \space + a_{k+1} b^{k+1} + a_k b^k$
$\space\space\space + a_{k-1}b^{k-1} + a_{k-2}b^{k-2} + \space ... \space + a_1 b + a_0$
We can rearrange the last expression by taking out powers of $b^k$ from each sequence of $k$ terms as follows
$\space\space\space + {\space a_{ck-1}b^{k-1} + a_{ck-2}b^{k-2} + \space ... \space + a_{(c-1)k+1} b + a_{(c-1)k} \space}.b^{(c-1)k}$
$\space\space\space + {\space a_{(c-1)k-1}b^{k-1} + a_{(c-1)k-2}b^{k-2} + \space ... \space + a_{(c-2)k+1} b + a_{(c-2)k} \space}. 532