UPDATE

I misunderstood your request. What you're asking for can be a giant pain to troubleshoot. Therefore, my brain blocked out your real question, and replaced it with something very close syntactically.
Now, I don't know if you know dominus, but he's usually a pretty mellow guy. Very helpful, always cheerful, & etc.. However, allow me to quote from the article myocom showed you :

It turned out that the clods who had written this program had done something like this:

        while (<RECORDS>) {
          chomp;
          @values = split /\t/, $_;
          foreach $v (@values) {
            $$v++;
          }
        }
[download]

you'll see that this bears a startling resemblance to what you've thought of. Don't do it.

Instead, he suggests

If the original programmers had used a series of hashes instead of stuffing everything into a bunch of global variables, it would never have happened, or at worst it would have been easy to fix. I ended up doing a major overhaul on the program to solve the problem.

Which is what I thought you needed help on. OK? OK.

now, back to the original post

---

here's one way :

for (a..z) {$in {$_}=int rand(100) };
foreach (keys %in){
    print "in $_ is $in{$_}\n";
    $$out{$_}=$in{$_}
}
foreach (keys %$out){print "out $_ is $$out{$_}\n"}
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the magic in line 4 works like this :
$out is normally a simple scalar. however, prepending it with another "$" and tacking on the {$_} makes $out a reference to a hash.
That's why we can use %$out in line 6. Using the "%" dereferences $out.
Similarly if we say  $foo=\$bar, $foo is holding a reference to that scalar, it's not a scalar in itself. You can dereference it with $$foo. Also,  $foo=\$bar can be written like  $$foo=$bar, which means "the scalar referenced by $foo set to $bar's value". Make sense?

When you do

foreach $field(@fieldList){
     $$field = $in{'$field'};
}
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, what you're doing is running through each element in @fieldlist as $field, and overwriting it with a reference to a scalar that's set to $in{'$field'}. Additionally, this doesn't make any sense either, since single quotes never do interpolation. Use double quotes or just get rid of them altogether, like $in{$field}.
Also, use strict and -w.
Finally, what reasons do you have for this reassignation of values? Will you really be transforming each $in variable? Or are you just wanting a temporary place to work with them?

If it's the later, dispense with this idea altogether, and just use %in in place.

Update : also, see tye's nifty-keen explanation of referencing and dereferencing for more details.