"\n" =~ /\n.*\z/;   # matches
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Obvious, I think. You match a "\n", then EOS (end of string).

"\n" =~ /.*\z/;     # doesn't match. i would expect it to match
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perl -D512 tells anchored(MBOL) (i.e. multiline beginning of line, see perldebguts) with that one, which anchoring doesn't happen with

"\n" =~ /[^\n]*\z/; # matches. like expected. but [\n]* is like .*
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but why?

"\n" =~ /.?\z/;
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matches, as does

"\n" =~ /.{0,}\z/;
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I can't get a mental model of why the previous one should, but the next one should not match:

"f\n" =~ /.?f\z/;
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Weird. Rather inconsistent, if not buggy.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}