There is nothing wrong with the idea of modifying an element inside a list while traversing it (don't modify the list itself, that's not recommended), but it can be better expressed like this:

for (int i = 0; i < letters.size(); i++) {
    letters.set(i, "D");
}

At the end the whole list will have the letter "D" as its content. It's not a good idea to use an enhanced for loop in this case, you're not using the iteration variable for anything, and besides you can't modify the list's contents using the iteration variable.

Notice that the above snippet is not modifying the list's structure - meaning: no elements are added or removed and the lists' size remains constant. Simply replacing one element by another doesn't count as a structural modification. Here's the link to the documentation quoted by @ZouZou in the comments, it states that:

A structural modification is any operation that adds or deletes one or more elements, or explicitly resizes the backing array; merely setting the value of an element is not a structural modification

ListIterator.set as returned by List.listIterator() or List.listIterator(int)

(set wouldn't make any sense for, say, a Set iterator.)

There is a built in method to do exactly this operation:

Collections.replaceAll(list, foo, bar);

The second version would be better. Internally they are the same in the end, but the second actually allows you to modify the list, while the first one will throw a ConcurrentModificationException.

But then you are using the Iterator in a wrong way. Here is how you do it correctly:

for (final ListIterator<String> i = list.listIterator(); i.hasNext();) {
  final String element = i.next();
  i.set(element + "yaddayadda");
}

The iterator is the one that needs to modify the list as it is the only one that knows how to do that properly without getting confused about the list elements and order.

Edit: Because I see this in all comments and the other answers:

Why you should not use list.get, list.set and list.size in a loop

There are many collections in the Java collections framework, each on optimized for specific needs. Many people use the ArrayList, which internally uses an array. This is fine as long as the amount of elements does not change much over time and has the special benefit that get, set and size are constant time operations on this specific type of list.

There are however other list types, where this is not true. For example if you have a list that constantly grows and/or shrinks, it is much better to use a LinkedList, because in contrast to the ArrayList add(element) is a constant time operation, but add(index, element), get(index) and remove(index) are not!

To get the position of the specific index, the list needs to be traversed from the first/last till the specific element is found. So if you do that in a loop, this is equal to the following pseudo-code:

for (int index = 0; index < list.size(); ++index) {
  Element e = get( (for(int i = 0; i < size; ++i) { if (i == index) return element; else element = nextElement(); }) );
}

The Iterator is an abstract way to traverse a list and therefore it can ensure that the traversal is done in an optimal way for each list. Test show that there is little time difference between using an iterator and get(i) for an ArrayList, but a huge time difference (in favor for the iterator) on a LinkedList.

The list is maintaining an object reference to the original value stored in the list. So when you execute this line:

Integer x = i.next();

Both x and the list are storing a reference to the same object. However, when you execute:

x = Integer.valueOf(9);

nothing has changed in the list, but x is now storing a reference to a different object. The list has not changed. You need to use some of the list manipulation methods, such as

list.set(index, Integer.valueof(9))

Note: this has nothing to do with the immutability of Integer, as others are suggesting. This is just basic Java object reference behaviour.

Here's a complete example, to help explain the point. Note that this makes use of the ListIterator class, which supports removing/setting items mid-iteration:

import java.util.*;

public class ListExample {

  public static void main(String[] args) {

    List<Foo> fooList = new ArrayList<Foo>();
    for (int i = 0; i < 9; i++)
      fooList.add(new Foo(i, i));

    // Standard iterator sufficient for altering elements
    Iterator<Foo> iterator = fooList.iterator();

    if (iterator.hasNext()) {
      Foo foo = iterator.next();
      foo.x = 99;
      foo.y = 42;
    }

    printList(fooList);    

    // List iterator needed for replacing elements
    ListIterator<Foo> listIterator = fooList.listIterator();

    if (listIterator.hasNext()) {
      // Need to call next, before set.
      listIterator.next();
      // Replace item returned from next()
      listIterator.set(new Foo(99,99));
    }

    printList(fooList);
  }

  private static void printList(List<?> list) {
    Iterator<?> iterator = list.iterator();
    while (iterator.hasNext()) {
      System.out.print(iterator.next());
    }
    System.out.println();
  }

  private static class Foo {
    int x;
    int y;

    Foo(int x, int y) {
      this.x = x;
      this.y = y;
    }

    @Override
    public String toString() {
      return String.format("[%d, %d]", x, y);
    }
  }
}

This will print:

[99, 42][1, 1][2, 2][3, 3][4, 4][5, 5][6, 6][7, 7][8, 8]
[99, 99][1, 1][2, 2][3, 3][4, 4][5, 5][6, 6][7, 7][8, 8]

Use the set method to replace the old value with a new one.

list.set( 2, "New" );

To replace the elements in-place without building a new list, you would need to reassign the elements at each index, that is, essentially zorglubs[i] = zorglubs[i].transform() for each index i in the list.

A perhaps elegant way to do this is using the eachWithIndex function, which gives you access to each element and its index:

zorglubs.eachWithIndex { it, index -> zorglubs[index] = it.transform() }