Does List<T> guarantee insertion order?

The List<> class does guarantee ordering - things will be retained in the list in the order you add them, including duplicates, unless you explicitly sort the list.

According to MSDN:

...List "Represents a strongly typed list of objects that can be accessed by index."

The index values must remain reliable for this to be accurate. Therefore the order is guaranteed.

You might be getting odd results from your code if you're moving the item later in the list, as your Remove() will move all of the other items down one place before the call to Insert().

Can you boil your code down to something small enough to post?

Here are 4 items, with their index

0  1  2  3
K  C  A  E

You want to move K to between A and E -- you might think position 3. You have be careful about your indexing here, because after the remove, all the indexes get updated.

So you remove item 0 first, leaving

0  1  2
C  A  E

Then you insert at 3

0  1  2  3
C  A  E  K

To get the correct result, you should have used index 2. To make things consistent, you will need to send to (indexToMoveTo-1) if indexToMoveTo > indexToMove, e.g.

bool moveUp = (listInstance.IndexOf(itemToMoveTo) > indexToMove);
listInstance.Remove(itemToMove);
listInstance.Insert(indexToMoveTo, moveUp ? (itemToMoveTo - 1) : itemToMoveTo);

This may be related to your problem. Note my code is untested!

EDIT: Alternatively, you could Sort with a custom comparer (IComparer) if that's applicable to your situation.

As Bevan said, but keep in mind, that the list-index is 0-based. If you want to move an element to the front of the list, you have to insert it at index 0 (not 1 as shown in your example).