Do I have a twin with permutated remainders?
R, 63 59 bytes



s=scan()%%c(2,3,5,7);i=which(s&lt;c(0,2,3,5));any(s[i]-s[i-1])


Try it online!

-4 bytes thanks to Giuseppe

(The explanation contains a spoiler as to how to solve the problem without computing \$k\$.)

Explanation:
 Let \$s\$ be the list of remainders. Note the constraint that s[1]&lt;2, s[2]&lt;3, s[3]&lt;5 and s[4]&lt;7. By the Chinese Remainder Theorem, there exists a \$k\$ iff there is a permutation of \$s\$, distinct from \$s\$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:


s[2]&lt;2 and s[2]!=s[1]
s[3]&lt;3 and s[3]!=s[2]
s[4]&lt;5 and s[4]!=s[3]