Everything Unique $II$

Obviously coding is the simplest way to go here, but it's just about possible manually.

Some initial observations:

1) $\overline{ABCD}$ is at least $1234$

2) $\overline{FGHI}$ is at most $9876$

3) $E$ can't be $1$ (because then $\overline{ABCD}$ would equal $\overline{FGHI}$ ) and $E$ can't be $9$ (because $9 \times 1234$ is too big)

4) $D$ can't be $5$ , because that would force $I$ to be $0$ or $5$ , neither of which is allowed

Now consider the problem modulo $9$ . If $A+B+C+D \equiv k$ , then $F+G+H+I \equiv kE$ . But the total of all the digits in the sum is $45$ ; ie $A+B+C+D+E+F+G+H+I \equiv 0 \pmod9$ so $\begin{aligned} k+E+kE &\equiv 0\pmod9 \\ (k+1)(E+1) &\equiv 1 \pmod9 \end{aligned}$

Remembering $1<E<9$ , this has solutions

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Case $E=7$ : $A+B+C+D \equiv 7 \pmod9$ . Since $A,B,C,D$ are distinct, the total can't be $7$ . Since $E=7$ , $\overline{ABCD}$ is at most $1428$ ; so $A=1$ and $B \le 4$ . Hence the total $A+B+C+D$ is less than $25$ ; so $A+B+C+D=16$ .

Remembering $D \neq 5$ , the only possible $\overline{ABCD}$ values this leaves are $\{1249,1258,1294,1348,1384$ and it's easy to check that none of these work.

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Case $E=6$ : In this case, $D$ cannot be even, as if it were then we would have $I=D$ . $A+B+C+D \equiv 3 \pmod9$ , and by its size, $\overline{ABCD} \le 1659$ . But $B$ can't be $6$ , so again we have $A=1$ and this time $B\le 5$ .

The possible totals of $A+B+C+D$ are then $12$ and $21$ . The possible values of $\overline{ABCD}$ this time are $1389,1479,1497,1587$ ; again it's easy enough to check that these don't work.

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Case $E=3$ : Now by its size, $\overline{ABCD} \le 3298$ , but of course we can't have $A=3$ . However, even with these restrictions, there are $16$ possibilities to check, which is a bit onerous. It turns out none of them work but it would be better to reduce these more - any ideas?

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Case $E=4$ : Finally, applying the same ideas, we find two solutions: namely $1738 \times 4=6952$ and $1963 \times 4=7852$

so $E=\boxed4$ .