If I want to take one "row" of my array-of-arrays and copy it to a separate array, is this correct: @SeparateArray = @{ MyAoA[$i] };

Barring the typo(?), yes that's right:

@SeparateArray = @{ $MyAoA[$i] };
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And if I want to do the opposite (copy my separate array into a row of my array-of-arrays), is this correct: @{ $MyAoA[$i] } = [ @SeparateArray ];

No, the types of either side of = must be, erm, equal. The first example below will add a reference to @SeparateArray to your AoA. That is, if you later change @SeparateArray, it'll change the contents of your AoA. Use the second example below if you don't want that to happen.

$MyAoA[$i] = \@SeparateArray; # reference
@{ $MyAoA[$i] } = @SeparateArray; # copy
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Finally, if I want to copy an entire array-of-arrays (and not just have references to the first one), can I simply write: @NewAoA = [ @OldAoA ]; or do I need to build a "for" loop to replace each component array separately, e.g.: @{ $NewAoA[$i] } = @{ $OldAoA[$i] };

Yes, you need a for loop to make copies of each of the array references.

Have you seen the Data Structures Cookbook?

update: fixed some typos
update^2: Added link to perldsc