Yes, you caught the general ideas, but I'll explain the parts you had questions on. Your formating of the code is correct, and the $a assignment is definatly there to throw you, which it didn't.

*$=sub{}; is equivalent to &$=sub{}. This is equivalent to sub $ {}, except that you can't do that. The magic is in the typeglob, which recognizes that its getting an anonymous sub, and names it &$. The value of $$ (the PID) is left unchanged. :-)

Next, you said: sub u(){$_} is the same as sub u { return $_; }, which it isn't. The parens() act as an empty prototype basically telling Perl that the letter u by itself is a valid function call. Otherwise you have to use &u or u(). This allows the u=>u in the next line. Also note that even though this sub is defined inside the scope of the other sub, it is still a package global subroutine - all subroutines are, regardless of where they are defined. Granted there is some magic involved with localized variables and closures, but this doesn't run into any of that.

s[]{82G5S="!A;F]T:&5R(%!E<FP\@:&%C:V5R}six;
[download]

Is kind of cool in that it takes $_, which is undefined, and replaces it with the payload, uuencoded. The six are just useless extensions, but distracting ones. Its also sort of a joke in that this was my sixth obfuscation posted here. Note that this will generate a warning about using an undefined variable, but will still pass strict.

And last but not least, die &$=>$/ calls the sub defined earlier, which basically does die unpack( u=>u), $/ Now you asked about $/, it's default value is "\n". If you die with a string ending in a newline, then the "at blah.pl line X" is supressed. That's also true with warn. The u=>u is equivalent to "u", u() where the first u tells unpack to uudecode the return from u(), which just returns $_.

Good job of taking this obfu apart.