Let $p$ be a prime number such that its index modulo $10$ be equal to $p-1$ (i.e. the numbers $0$, $1$, $10$, $\ldots$, $10^{p-2}$ form a complete residue system modulo $p$.) Let $N(p)$ be the number $\frac{10^{p-1}-1}{p}$ with added leading zeroes in order to be a $(p-1)$-digit number. Then the numbers $iN(p)$ for $1 \le i \le p-1$, each one with added leading zeroes in order to be a $(p-1)$-digit number, are anagrams of $N(p)$.

Indeed for $1 \le i \le p-1$ we can find $k$ such that $10^k \equiv i \pmod p$. Now $iN(p)$ is a period of the repeating decimal $\frac{i}{p}$, and hence of $\frac{10^k}{p}$. However the period of the latter is a cyclic permutation of the period of $\frac{1}{p}$, which equals $N(p)$.

The index of $17$ modulo $10$ is $16$. If leading zeroes were allowed, the numbers $N(17)$, $2N(17)$, $3N(17)$ and $4N(17)$ would provide a suitable example. To bypass the leading zeroes problem, we can glue to the left of each of these numbers a number $s$ with no leading zeroes which is an anagram of $3s$. Since the index of $7$ modulo $10$ is $6$ and $7 < 10$, a suitable example is $N(7) = 142857$. Thus a possible example is given by the numbers $10^{16}N(7) + iN(17)$ for $1 \le i \le 4$.