They are kind of hard to come up with, but okay.

Given the rep 'aaaabaaaaba' and a string containing one whole and one partial rep 'aaaabaaaabaaaaabaaaab'

$s = 'aaaabaaaabaaaaabaaaab';;
$s =~ m[(.+).*\1] and print $1;;
aaaabaaaab
[download]

Which isn't correct because:

aaaabaaaabaaaaabaaaab
aaaabaaaab aaaabaaaab
1         ^2
[download]

You can fix that by removing the redundant .* per LanX's version: m[(.+)\1] but then you get:

$s = 'aaaabaaaabaaaaabaaaab';;
$s =~ m[(.+)\1] and print $1;;
aaaab
[download]

Which isn't right:

aaaabaaaabaaaaabaaaab
aaaabaaaab aaaabaaaab
1    2    ^3    4
[download]

I realise that this is a 'cheat' as there in no complete repetition to find, but it is one possible scenario.

Given the string will always consist of 1 or more repetitions of the substring, whatever partial substring (if any) is at the end of the string should match the same number of characters at the start of the string. That's the bit I'm having trouble wrapping my head around.