Does the isbn detect jump transpositions?
$$a_1+2a_2+3a_3+\cdots+10a_{10}=0\pmod{11}$$
I think it does because the specific formula multiplying 1 times 1st digit, 2 times 2nd digit... will give you a remainder of zero if the isbn is correct. if it's not the right isbn, then you will get a remainder $\ge 1$.
Call the sum $a_1+2a_2+3a_3+\cdots+10a_{10}=S$. We have $S\equiv 0\pmod{11}$. Suppose we transpose the $i^\text{th}$ and $j^\text{th}$ digits. In order to not detect this transposition, we must have $$S-ia_i+ia_j-ja_j+ja_i\equiv 0\pmod{11}$$ We now remove $S$ (which is 0), and rearrange to get $$(j-i)a_i\equiv (j-i)a_j\pmod{11}$$
We now need one of the important properties of modular arithmetic:
If $ac\equiv bc\pmod{n}$ and $\gcd(c,n)=1$, then $a\equiv b\pmod{n}$
Since $11$ is prime, $i\neq j$, and $1\le i,j\le 10$, we know $\gcd(j-i,11)=1$. Hence we may conclude that $a_i\equiv a_j\pmod{11}$.
Hence the only transpositions we can't detect are of the same two digits transposed anyway.