The number $(12\ldots(b-1))$ in base $b$ has the property that when multiplied by any integer $1\le k\le b-1$ which is coprime to $b-1$, its digits are permuted. Why?
For example in base 10, \begin{eqnarray} 123456789&*2=& 246913578\\ &*4=& 493827156\\ &*5=& 617283945\\ &*7=& 864197523\\ &*8=& 987654312 \end{eqnarray}
I'll rederive Henry's result, and extend it, since I did it while he entered his comment.
If we replicate the OP's computation, but with multiplier 3, we find that 3 x 123456789 = 370370370370.
This leads to the conjecture that if $m$ and $b-1$ are relatively prime, then $m\cdot s_b$ is a permutation of the digits.
I can show that this is true for $m = b-2$. The method I use might be generalized for other $m$.
(Comment added later: if we look at the products for base 10, notice that for 8 there are 8 decreasing digits before the final 2, for 7 there are 4 and 4 decreasing before the final 3, and for 5 there are 4 pairs of decreasing digits before the final 5. This might be provable using the results at the end of my answer.)
To show $s_b =\sum_{k=1}^{b-1} k b^{b-k-1} =\frac{b^b-b}{(b-1)^2}-1 $.
Since
$\begin{array}\\ \sum_{k=1}^{n} k r^{k-1} &=\frac{1-r^{n+1}}{(1-r)^2}-\frac{(n+1)r^n}{1-r}\\ &=\frac{1-r^{n+1}-(1-r)(n+1)r^n}{(1-r)^2}\\ &=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\\ \end{array} $
putting $r = 1/b$ and $n = b-1$,
$\begin{array}\\ s_b &=\sum_{k=1}^{b-1} k b^{b-k-1}\\ &=b^{b-2}\sum_{k=1}^{b-1} k b^{-k+1}\\ &=b^{b-2}\sum_{k=1}^{b-1} k (1/b)^{k-1}\\ &=b^{b-2}\frac{1-b(1/b)^{b-1}+(b-1)(1/b)^b}{(1-1/b)^2}\\ &=\frac{b^{b-2}-1+(b-1)/b^2}{(1-1/b)^2}\\ &=\frac{b^b-b^2+b-1}{(b-1)^2}\\ &=\frac{b^b-(b^2-b+1)}{(b-1)^2}\\ &=\frac{b^b-b-(b^2-2b+1)}{(b-1)^2}\\ &=\frac{b^b-b}{(b-1)^2}-1\\ &=b\frac{b^{b-1}-1}{(b-1)^2}-1\\ &=b\frac{(b-1)\sum_{k=0}^{b-2}b^k}{(b-1)^2}-1 \quad\text{(now I try to look at the base b digits)}\\ &=b\frac{\sum_{k=0}^{b-2}b^k}{b-1}-1\\ &=\frac{\sum_{k=0}^{b-2}b\ b^k}{b-1}-1\\ &=\frac{\sum_{k=0}^{b-2}(b-1+1)\ b^k}{b-1}-1\\ &=\sum_{k=0}^{b-2} b^k+\frac{\sum_{k=0}^{b-2}b^k}{b-1}-1\\ &=\sum_{k=1}^{b-2} b^k+\frac{\sum_{k=0}^{b-2}b^k}{b-1}\\ \end{array} $
From this, $m\cdot s_b =m\sum_{k=1}^{b-2} b^k+m\frac{\sum_{k=0}^{b-2}b^k}{b-1} $. This is, in base $b$, $b-2$ $m$'s and a zero digit added to $m$ times $b-1$ $m$'s divided by $b-1$.
If $m=b-2$, which is relatively prime to $b-1$,
$\begin{array}\\ (b-2)\cdot s_b &=(b-2)\sum_{k=1}^{b-2} b^k+(b-2)\frac{\sum_{k=0}^{b-2}b^k}{b-1}\\ &=(b-2)\sum_{k=1}^{b-2} b^k+(b-1-1)\frac{\sum_{k=0}^{b-2}b^k}{b-1}\\ &=(b-2)\sum_{k=1}^{b-2} b^k+\sum_{k=0}^{b-2}b^k-\frac{\sum_{k=0}^{b-2}b^k}{b-1}\\ &=1+(b-1)\sum_{k=1}^{b-2} b^k-\frac{\sum_{k=0}^{b-2}b^k}{b-1}\\ &=(b-2)+(b-1)\sum_{k=1}^{b-2} b^k-\frac{b^{b-1}-1}{(b-1)^2}\\ &=1+(b-1)\sum_{k=1}^{b-2} b^k-\frac{s_b+1}{b}\\ \end{array} $.
$s_b+1 =1+\sum_{k=1}^{b-1} k b^{b-k-1} =b+\sum_{k=1}^{b-2} k b^{b-k-1} $ so $\frac{s_b+1}{b} =1+\sum_{k=1}^{b-2} k b^{b-k-2} $ and this has, in base $b$, the digits $1, 2, ..., b-4, b-3, b-1$ (the 1 changes the last b-2 to b-1).
Subtracting this from $b-1, b-1, ..., b-1, b-1, 1$, we get $b-1, b-2, b-3, ..., 3, 1, 2$ which is a permutation of 1 through $b-1$.
That's enough for me, for now.
