Kotlin merge two nullable mutable list

Here are couple of solutions.

1. In case if you need to add all elements to mutableList1:

   val mutableList1: MutableList<Any?>? = ...
   val mutableList2: MutableList<Any?>? = ...
   
   mutableList1?.let { list1 -> mutableList2?.let(list1::addAll) }
   

2. In case if you need new nullable list as result:

   val mutableList1: MutableList<Any?>? = ...
   val mutableList2: MutableList<Any?>? = ...
   
   val list3: List<Any?>? = mutableList1?.let { list1 ->
       mutableList2?.let { list2 -> list1 + list2 }
   }
   

3. In case if you need new nullable mutable list as result:

   val mutableList1: MutableList<Any?>? = ...
   val mutableList2: MutableList<Any?>? = ...
   
   val list3: MutableList<Any?>? = mutableList1
           ?.let { list1 -> mutableList2?.let { list2 -> list1 + list2 } }
           ?.toMutableList()
   

4. In case if you need new non-null list as result:

   val mutableList1: MutableList<Any?>? = ...
   val mutableList2: MutableList<Any?>? = ...
   
   val list3: List<Any?> = mutableList1.orEmpty() + mutableList2.orEmpty()
   

Based on your snippets, which don't make a consistent whole, I made some guesses as to what you actually wanted to achieve:

val mutableList1: MutableList<String?>? = ...
val mutableList2: MutableList<String?>? = ...

val mapped1 = mutableList1?.mapTo(ArrayList()) { TeamInvitationData(it) }
val mapped2 = mutableList2?.mapTo(ArrayList()) { TeamInvitationData(it) }

mapped1?.addAll(mapped2.orEmpty())

The key point to note is that map() returns an immutable list regardless of the type of the input list. To get a mutable list you must use mapTo(destination) { ... }. Once that is fixed, you can use addAll() as shown in the last line.

plus. Returns an List containing all elements of the original collection and then the given Iterable. source

Collection<T>.plus(elements: Iterable<T>): List<T>

Another Good read here