Try:

perl -le 'print $ARGV[-1] =~ s/[\da-zA-Z]+(?=\.)/++($i=$&)/er' file*.txt 

That will give you file_10.txt after file_9.txt, file_g.txt after file_f.txt, file_aa.txt after file_z.txt, but not file_ab.txt after file_aa.txt or file_11.txt after file_10.txt because the file* shell glob will sort file_z.txt after file_aa.txt and file_9.txt after file_10.txt (that latter one you can work around with zsh by using file*.txt(n) instead of file*.txt).

Try:

perl -le 'print $ARGV[-1] =~ s/[\da-zA-Z]+(?=\.)/++($i=$&)/er' file*.txt 

That will give you file_10.txt after file_9.txt, file_g.txt after file_f.txt, file_aa.txt after file_z.txt, but not file_ab.txt after file_aa.txt or file_11.txt after file_10.txt because the file* shell glob will sort file_z.txt after file_aa.txt and file_9.txt after file_10.txt.

That latter one you can work around with zsh by using file*.txt(n) instead of file*.txt.

Or you can define a numeric sort order in zsh, based on those aa, abc being recognised as numbers in base 36:

b36() REPLY=$((36#${${REPLY:r}#*_})) perl ... file_*.txt(no+b36) 

(note that the order is ...7, 8, 9, a/A, b/B..., z/Z, 10, 11... so you don't want to mix file_123.txt and file_aa.txt).

Try:

perl -le 'print $ARGV[-1] =~ s/[\d\a]+(?=\.)/++($i=$&)/er' file*.txt 

That will give you file_10.txt after file_9.txt, file_g.txt after file_f.txt, file_aa.txt after file_z.txt, but not file_ab.txt after file_aa.txt because the file* shell glob will sort file_z.txt after file_aa.txt.

Try:

perl -le 'print $ARGV[-1] =~ s/[\da-zA-Z]+(?=\.)/++($i=$&)/er' file*.txt 

That will give you file_10.txt after file_9.txt, file_g.txt after file_f.txt, file_aa.txt after file_z.txt, but not file_ab.txt after file_aa.txt or file_11.txt after file_10.txt because the file* shell glob will sort file_z.txt after file_aa.txt and file_9.txt after file_10.txt (that latter one you can work around with zsh by using file*.txt(n) instead of file*.txt).