For any choice of 6 distinct non-zero digits, at least two of them are consecutive.
Indeed, if $a < b < c < d < e < f$ and no digits are consecutive, then $b \ge a + 2$, $c \ge b + 2 \ge a + 4$, $d \ge c + 2 \ge a + 6$, $e \ge d + 2 \ge a + 8$ and $f \ge e + 2 \ge a + 10$, which is impossible, since both $a$ and $f$ are digits.
Denote by $m$ and $n$, $m > n$, two consecutive digits among $a, b, c, d, e, f$, and by $p, q, r, s$ the other four digits.
We have $x \mid \overline{pqrsmn}$ and $x \mid \overline{pqrsnm}$, hence $x \mid \overline{pqrsmn} - \overline{pqrsnm} = 9(m-n)$. Since $m-n=1$, we find $x \mid 9$.
If $3 \nmid a + b + c + d + e + f$, the only possibility is $x = 1$.
If $3 \mid a + b + c + d + e + f$ and $9 \nmid a + b + c + d + e + f$, then $x = 1$ or $x = 3$.
If $9 \mid a + b + c + d + e + f$, then $x = 1, x = 3$ or $x = 9$.