我发现它只需要一些调整：

///example: source = 1, destination = 4
///original list: [a,b,c,d,e]
///final list: [b,c,d,a,e]
func move(from source: IndexSet, to destination: Int)
         var sortedOrders: [Int16] = Array(1...Int16(myEntities.count))
        for (entity, order) in zip(myEntities, sortedOrders) {
            entity.order = order
        }
        
        sortedOrders.move(fromOffsets: source, toOffset: destination)
        var finalOrders : [Int16] = Array(1...Int16(myEntities.count))
        var mappingDict = Dictionary(uniqueKeysWithValues: zip(sortedOrders , finalOrders ) )
        // [5: 5, 4: 3, 2: 1, 1: 4, 3: 2]
        
        for entity in myEntities{
            entity.order = mappingDict[entity.order]!
        }

}

下面是我对@luomein的回答的解释，它提供了一种通用的方法来获得索引Map，使得move函数的实现简单得多。
首先，我们将这个扩展添加到Collection：

extension Collection {
    func newIndices(moving source: IndexSet, to destination: Int) -> [Int: Int] {
        var oldIndexesByNewIndex: [Int] = Array(0..<self.count)
        oldIndexesByNewIndex.move(fromOffsets: source, toOffset: destination)
        return Dictionary(uniqueKeysWithValues:
                            oldIndexesByNewIndex.enumerated().map{ newIndex, oldIndex in (oldIndex, newIndex) }
        )
    }
    
}

然后我们这样使用它：

func move(from source: IndexSet, to destination: Int) {
        let newIndices = entities.newIndices(moving: source, to: destination)
        entities.enumerated().forEach { currentIndex, entity in
            if (newIndices[currentIndex] != currentIndex) {
                entity.order = newIndices[currentIndex]!
            }
        }
    }