Copy a file after renaming another file

Please paste the code below into a standard code module in any blank workbook. Then run the procedure 'CopyAndRename'.

Option Explicit
Option Base 0

Sub CopyAndRename()

    ' modify path as required
    Const PathName As String = "C:\Users\abcuser\AppData\Local\App01"
    
    Dim SelFiles() As String
    Dim NewFn As String                     ' Fn = file name
    Dim OldFn As String
    
    If GetSelectedFiles(PathName, SelFiles) Then
        NewFn = NewFileName(SelFiles(0), OldFn)
        If Len(NewFn) Then
            FileCopy SelFiles(0), NewFn
            Name SelFiles(0) As OldFn
        Else
            MsgBox "The selected file has no exension.", _
                   vbCritical, "Invalid file name"
        End If
    End If
End Sub

Private Function GetSelectedFiles(Pn As String, _
                                  SelFiles() As String) _
                                  As Boolean
    ' 0086 V 1.0
    
    Dim FoD As FileDialog                           ' File Open Dialog
    Dim SelItem As Variant                          ' Selected Item
    Dim i As Long                                   ' Index for SelFiles
    
    Set FoD = Application.FileDialog(msoFileDialogFilePicker)
    With FoD
        .Title = "Choose the file to copy"
        .ButtonName = "Create Copy"
        .Filters.Clear
        .Filters.Add "XML files", "*.xml", 1
        .InitialFileName = WithSeparator(Pn)
        .AllowMultiSelect = False
        If .Show Then
            For Each SelItem In .SelectedItems
               ReDim Preserve SelFiles(i)
               SelFiles(i) = SelItem
               i = i + 1
            Next SelItem
            GetSelectedFiles = True
        End If
    End With
    Set FoD = Nothing
End Function

Private Function NewFileName(Ffn As String, _
                             OldFn As String) As String

    Dim Sp() As String
    Dim Fn As String                ' File name (original)
    Dim Ext As String
    
    Sp = Split(Ffn, ".")
    If UBound(Sp) Then
        Ext = Sp(UBound(Sp))
        Fn = Left(Ffn, Len(Ffn) - (Len(Ext) + 1))
        OldFn = Fn & "Old." & Ext
        NewFileName = Fn & "New." & Ext
    End If
End Function

Private Function WithSeparator(ByVal PathName As String, _
                               Optional ByVal RemoveExisting As Boolean) _
                               As String
                       
    Do While Right(PathName, 1) = Application.PathSeparator
        PathName = Left(PathName, Len(PathName) - 1)
    Loop
    WithSeparator = PathName & IIf(RemoveExisting, "", _
                                   Application.PathSeparator)
End Function

Select all Open in new window

The program will allow you to select a file, starting in the directory you mentioned (you can choose another default by assigning another string to the constant 'PathName' at the top of the code). When you press the "Create Copy" button in the file_open dialog box the selected file will be renamed as [file]New.[Ext] and a copy will be created as [file]Old.[Ext]
I wonder if this is your intention because no file of the original name will remain. However, if I misunderstood your instruction this is a small thing to correct in the above code which I shall be glad to do.