尝试以下操作以获取新文件名

Files = @( "Z:\Test\1\Script\Export_2022-11-09.csv", "Z:\Test\2\Script\Export_2022-11-09.csv")
foreach($File in $Files)
{
   $lastIndex = $File.LastIndexOf("\")
   $baseFilename = $File.Substring($lastIndex + 1)
   $newFilename = $File.Substring(0, $lastIndex + 1)
   $newFilename += "Archive\" + $baseFileName
   Write-Host "New File Name = " $newFilename
}

似乎您想要的是将原始文件保存在同一路径下的归档子文件夹中。然后重写原始文件，以便删除标题行。
使用[System.IO.Path]::GetDirectoryName()将路径从完整路径和文件名字符串中分离出来，然后用‘存档’将其连接起来

$Files = "Z:\Test\1\Script\Export_2022-11-09.csv", "Z:\Test\2\Script\Export_2022-11-09.csv"

$Files | ForEach-Object {
    # construct the archive path
    $archivePath = Join-Path -Path ([System.IO.Path]::GetDirectoryName($_)) -ChildPath 'Archive'
    # create that folder if it does not already exist
    $null = New-Item -Path $archivePath -ItemType Directory -Force
    # copy the file as-is
    Copy-Item -Path $_ -Destination $archivePath

    # now remove the header line from the original file
    (Get-Content $_ | Select-Object -Skip 1) | Set-Content $_

    # for the final part, Move to another folder where a batch reads them, I will need more info
    # does every file have the SAME final destination folder? In that case, since all files are
    # named the same you will get file naming collisions there..
}