Or in any base that’s a perfect square (except for the trivial example of 1/2 in some sense always working out).
The phenomenon we’re interested in is that, for each x not divisible by 7, the fractional component of x/7 matches that of 1/7 * a power of 10. In other words, in mod 7 land, the powers of 10 include all the nonzero values (i.e., all the values strictly between 0 and 7).
More generally, we may replace 7 by some m and our decimal base by base B, and ask whether this still works.
You can quickly convince yourself this would be impossible in any base if m weren’t prime (in mod m land, powers of values coprime to m remain coprime to m, while powers of values non-coprime to m remain non-coprime to m, so powers of B couldn’t include both; it’d have to be the case that every value not divisible by m were coprime to m, which is to say, that m is prime).
Now, m = 2 always works, but only trivially so (every multiple of 1/2 is either straight-up an integer or an integer + 1/2; there’s only one fractional component that ever comes up, so it’s not very interesting).
Every other prime m will be odd. And in that case, the m - 1 nonzero values mod m pair off into (m - 1)/2 values up to negation, producing (m - 1)/2 distinct square values, so not all of them will be squares mod m. Yet, if b is a square, then so must be every power of b; thus, square b will be unsuitable.
This explains why there’s no “cyclic fraction” like 1/7 for base 16 or any other square base (except in the trivial sense of 1/2).
[A nice addendum to the above is that EVERY prime m comes with at least 1 suitable b (indeed, precisely as many suitable b as there are values coprime to m - 1). This is one of the great hidden (to laypeople) treasures of mathematics, I think. I’ll phrase the argument for this a bit more jargonly: arithmetic modulo a prime forms a field, and thus, there are at most n values satisfying x^n = 1, for each n. Thus, its multiplicative group has at most n elements of order dividing n, for each n.
Any finite such group is cyclic by a simple counting argument: Any value of order precisely n induces, via its powers, all n elements of order dividing n, of which precisely phi(n) will have order precisely n; thus, there are either 0 or phi(n) elements of order precisely n. Summing this up over all the proper divisors of the size |G| of the group, it follows that there are left over at least (and thus exactly) phi(|G|) generators of the group.]