How can I represent a "many to many" relation with Android Room when column names are same?

With the introduction to Junction in room you can handle many-to-many relationship with ease.

As @Devrim stated you can use an extra entity (ReservationGuest) which keeps the relation between Guest and Reservation(also know as associative table or junction table or join table).

@Entity
data class Guest(
  @PrimaryKey
  val gId: Long,
  val name: String,
  val email: String
)

@Entity
data class Reservation(
  @PrimaryKey
  val rId: Long,
  val table: String
)

@Entity(
  primaryKeys = ["reservationId", "guestId"]
)
data class ReservationGuest(     
  val reservationId: Long,
  val guestId: Long
)

Now you can get reservation with guests using this model:

data class ReservationWithGuests (
    @Embedded
    val reservation: Reservation,
    @Relation(
            parentColumn = "rId",
            entity = Guest::class,
            entityColumn = "gId",
            associateBy = Junction(
                    value = ReservationGuest::class,
                    parentColumn = "reservationId",
                    entityColumn = "guestId"
            )
    )
    val guests: List<Guest>
)

You can also get guest with their list of reservations as.

data class GuestWithReservations (
  @Embedded
  val guest: Guest,
  @Relation(
        parentColumn = "gId",
        entity = Reservation::class,
        entityColumn = "rId",
        associateBy = Junction(
                value = ReservationGuest::class,
                parentColumn = "guestId",
                entityColumn = "reservationId"
        )
  )
  val reservations: List<Reservation>
)

Now you can query database for the result as:

@Dao
interface GuestReservationDao {
  @Query("SELECT * FROM Reservation")
  fun getReservationWithGuests(): LiveData<List<ReservationWithGuests>>

  @Query("SELECT * FROM Guest")
  fun getGuestWithReservations(): LiveData<List<GuestWithReservations>>

}

Actually there is one more possibility to get Guest list, not only id's like in @Devrim answer.

First define class which will represent the connection between Guest and Reservation.

@Entity(primaryKeys = ["reservationId", "guestId"],
        foreignKeys = [
            ForeignKey(entity = Reservation::class,
                    parentColumns = ["id"],
                    childColumns = ["reservationId"]),
            ForeignKey(entity = Guest::class,
                    parentColumns = ["id"],
                    childColumns = ["guestId"])
        ])
data class ReservationGuestJoin(
    val reservationId: Long,
    val guestId: Long
)

Each time you will be inserting new Reservation, you will have to insert ReservationGuestJoin object in order to fulfill foreign key constraint. And now if you want to get Guest list you can use power of SQL query:

@Dao
interface ReservationGuestJoinDao {

    @SuppressWarnings(RoomWarnings.CURSOR_MISMATCH)
    @Query("""
        SELECT * FROM guest INNER JOIN reservationGuestJoin ON
        guest.id = reservationGuestJoin.guestId WHERE
        reservationGuestJoin.reservationId = :reservationId
        """)
    fun getGuestsWithReservationId(reservationId: Long): List<Guest>
}

To see more details visit this blog.

Based on the answer above: https://stackoverflow.com/a/44428451/4992598 only by keeping separate field names between entities you can have models returned (not just ids). All you need to do is:

@Entity data class ReservationGuest(
    @PrimaryKey(autoGenerate = true) val id: Long,
    val reservationId: Long,
    @Embedded
    val guest: Guest
)

And yes entities can be embedded in one another as long as you don't keep duplicate fields. So in consequence the ReservationWithGuests class can look like this.

data class ReservationWithGuests(
    @Embedded val reservation:Reservation,
    @Relation(
        parentColumn = "id",
        entityColumn = "reservationId",
        entity = ReservationGuest::class,
        projection = "guestId"
    ) val guestList: List<Guest>
)

So at this point you can use val guestIdList: List because your ReservationGuest entity actually maps ids with entity models.

For the join table entity, I suggest to use a composite ID indexed:

@Entity(
    primaryKeys = ["reservationId", "guestId"],
    indices = [Index(value =["reservationId", "guestId"], unique = true)]
)
data class ReservationGuestJoin(
    @PrimaryKey(autoGenerate = true) var id: Long,
    var reservationId: Long = 0,
    var guestId: Long = 0
)

The GuestDao.kt:

@Dao
@TypeConverters(GuestDao.Converters::class)
interface GuestDao {

    @Query(QUERY_STRING)
    fun listWithReservations(): LiveData<List<GuestWithReservations>>

    data class GuestWithReservations(
        var id: Long? = null,
        var name: String? = null,
        var email: String? = null,
        var reservations: List<Reservation> = emptyList()
    )

    class Converters{
        @TypeConverter
        fun listReservationFromConcatString(value: String?): List<Reservation>? = value?.let { value ->
                .split("^^")
                .map { it.split("^_") }
                .map { Reservation(id = it.getOrNull(0)?.toLongOrNull(), name = it.getOrNull(1)) }
        } ?: emptyList()
    }
}

The QUERY_STRING. We make a inner joins to produce a big table with data from both entities, them we concatenate the data from Reservation as a column string and lastly we group_concat the rows by the guest ID, concatenating the reservation strings with different separators, our converter will take care of rebuild it as a entity:

SELECT 
    t.id, t.name, t.email, GROUP_CONCAT(t.reservation, '^^') as reservations 
FROM (
    SELECT 
        guestId as id, name, email, (reservationId || '^_' || reservationTable) as reservation 
    FROM  
        GuestReservationJoin
        INNER JOIN Guest ON Guest.id = GuestReservationJoin.guestId 
        INNER JOIN Reservation ON Reservation.id = GuestReservationJoin.reservationId
    ) as t 
GROUP BY t.id

Note that I changed your column table name because I think Room do not allow you to use SQLite reserved names.

I didn't test the performance of all this compared with have more flat entity (another option without the concatenations). If I do, I'll update my answer.