Is there something similar for replacing, something like...

my @strs = qw/fooxx baryyyy bazzzzzzz/;
s{ ( (.) \2+ ) \z }(substr($1, 0, length($1) / 2))ex, print
  for @strs;

__output__

foox
baryy
bazaaa
[download]

In case you're working on large strings the above regex would be quite hefty in terms of processing, so here's a version that would perform much better on longer strings

reverse =~ /\A((.)\2+)/ and substr($_, -(length($1) / 2)) = ''
  for @strs;
print map "$_\n", @strs;

__output__

foox
baryy
bazaaa
[download]

_________
broquaint

update: added second code example

In a regular expression, you can use variables as well: $variable =~ s/($pattern)+/..../ Beware that $pattern is interpolated, so if $pattern contains something like \, it may even become an error, e.g

C:\>perl
$pattern = "\\";
$variable = "abcde\\fg";
print $variable =~ /$pattern/;
^D
Trailing \ in regex m/\/ at - line 4.
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$variable =~ s/
  (  # start capturing to $1
    (?: # not capturing, just clustering
      \Q$pattern\E # $pattern in a quoted way
    )
    + # one or more times
  )  # end capturing to $1
/
....
/x;
[download]

  $variable =~ s/((?:\Q$pattern\E)+)/$1.$1.$1/e;
  print $variable;
[download]

Best regards,
perl -e "s>>*F>e=>y)\*martinF)stronat)=>print,print v8.8.8.32.11.32"

s/((.)\2*)\1$/$1/;
[download]

This wasn't tested, I assume, as it doesn't work as requested. Try it with my string below.

Update: To clarify: 'aaabcdabcddddddefgg' would become 'aaabcdabcddddddefg' using your version.

Update 2: Oops. I misread the question as 'all' duplicates in the string. Thanks to physi for pointing out my fault. Interesting exercise, anyway.

Update: as stated before, I misread the question. There is no problem with Skeeve's solution, as it does what is stated.

Because I couldn't get Skeeve's to work on my strings, I did my own version, which should work:

my $string = 'aaabcdabcddddddefgg';
$string =~ s/((.)(?(?=\2{2,})\2+))\1/$1/g;
print "$string\n";
[download]

I used the 'if 2 or more of the previous string' swallow them (prior to the backtrack, but anyhow), as it wouldn't work with duplicates when two or three were back to back, and * wasn't the solution, either.

Also to note, this 'rounds up', so if there are three back to back, two will remain (3 / 2 = 1.5, rounded up to 2).

As I understand the question, it was only for the pattern at the end of the string. So Skeeve's working very well.
If you want to half every pattern in the string, just do:

s/((.)\2*)\1/$1/g;
[download]

That gives the same output like your does.

-----------------------------------
--the good, the bad and the physi--
-----------------------------------

Quite complicated, isn't it? As physi already stated, a simple:

s/((.)\2*)\1/$1/g
[download]

will do the trick of half-ing each repeated occurence and rounding them up. So "aaabbbb" will become "aabb".