I'm actually working on something similar and ran into the same issues. The solution I came up with was to use the File option (not the UploadFile one), write the input to another file, then perform the unzip.

Here's my implementation using a TemporaryDirectory and separate directories for input and unzipped files:

@router.post("/test_file/unzip")
def unzip_upload(file: bytes = File(...)):

    with tempfile.TemporaryDirectory() as temp_dir:
        os.chdir(temp_dir)
        os.mkdir('input')
        os.mkdir('unzipped')

        with open("input/zip_file.zip", 'wb') as new_file:
            new_file.write(file)
            # print(f"output of listdir for /input {os.listdir(temp_dir + '/input')}")

            with zipfile.ZipFile("input/zip_file.zip") as zip_file:
                print(f"files in zip: {zip_file.namelist()}")
                zip_file.extractall('unzipped')

        unzipped_files = os.listdir('unzipped')

    return {"unzipped files": unzipped_files}

If you need to use the UploadFile option, I got this to work:

@router.post("/test_uploadfile/unzip")
def unzip_upload(file: UploadFile = File(...)):

    with tempfile.TemporaryDirectory() as temp_dir:
        os.chdir(temp_dir)
        os.mkdir('input')
        os.mkdir('unzipped')

        with open("input/zip_file.zip", 'wb') as new_file:
            new_file.write(file.file._file.getvalue())
            # print(f"output of listdir for /input {os.listdir(temp_dir + '/input')}")

            with zipfile.ZipFile("input/zip_file.zip") as zip_file:
                print(f"files in zip: {zip_file.namelist()}")
                zip_file.extractall('unzipped')

        unzipped_files = os.listdir('unzipped')

    return {"unzipped files": unzipped_files}

@app.post("/", status_code=200)
async def upload(file: bytes = File(...)):
 zip_file_name = "file.zip"
    
 with open(zip_file_name, 'wb') as zip_file:
  zip_file.write(file)

I switched from accepting a tempfile.TemporaryDirectory() to a bytes as suggested by C H here.