You have 3 options here:

Using third-party packages which provide menu

This is pretty straight forward, there are some good packages out there which support menu for admin, and some way to add your item to the menu.

An example of a 3rd party package would be django-admin-tools (last updated Dec 2021, checked April 2023). The drawback is that it is a bit hard to learn. Another option would be to use django-adminplus (last updated Oct 2019, checked April 2023) but at the time of writing this, it does not support Django 2.2.

Hacking with django admin templates

You can always extend admin templates and override the parts you want, as mentioned in @Sardorbek answer, you can copy some parts from admin template and change them the way you want.

Using custom admin site

If your views are supposed to only action on admin site, then you can extend adminsite (and maybe change the default), beside from adding links to template, you should use this method to define your admin-only views as it's easier to check for permissions this way.

Considering you already extended adminsite, now you can use the previous methods to add your link to template, or even extend get_app_list method, example:

class MyAdminSite(admin.AdminSite):
    def get_app_list(self, request):
        app_list = super().get_app_list(request)
        app_list += [
            {
                "name": "My Custom App",
                "app_label": "my_test_app",
                # "app_url": "/admin/test_view",
                "models": [
                    {
                        "name": "tcptraceroute",
                        "object_name": "tcptraceroute",
                        "admin_url": "/admin/test_view",
                        "view_only": True,
                    }
                ],
            }
        ]
        return app_list

You can also check if current staff user can access to module before you show them the links. It will look like this at then end:

This guide looks quite old. I would rather advise you to follow django docs.

someapp/admin.py

from django.contrib.admin import AdminSite
from django.http import HttpResponse

class MyAdminSite(AdminSite):

     def get_urls(self):
         from django.urls import path
         urls = super().get_urls()
         urls += [
             path('my_view/', self.admin_view(self.my_view))
         ]
         return urls

     def my_view(self, request):
         return HttpResponse("Hello!")

admin_site = MyAdminSite()

Source: https://github.com/django/django/blob/2.2/django/contrib/admin/sites.py#L194-L205

You should also update your project/urls.py and replace path('admin/', admin.site.urls) by path('admin/', admin_site.urls). Don't forget to from someapp.admin import admin_site before.

Rather then creating a custom admin change view I would approach this problem by extending the built in one, the django admin provides a number of hooks to facilitate this, here are my thoughts:

1. Modify the default queryset but I would recommend against this as you loose the ability to edit objects not returned by the queryset.
2. Create a custom filter that would display your 'broker list' queryset - I recommend this one as it further will give you a GET url query you can link to directly to activate this filter.
3. You can even completely gut out the change view and use your own - this option I have the least experience with and can not comment.

You may wish to also take at look changing the change list template and admin custom actions to further customize the look and feel and to provide custom 'actions'; I have used both successfully in the past to provide project specific functionality.

Very exciting are the has_add_premission, had_change_permission, and has_delete_permission hooks, couple these with like django-guradian and custom admin base template could allow you to use the backend admin as a complete front end administration.

Do take the time to throughly read the entire model admin page - the better I know it the less I find myself coding custom front end administration.

It is solved. I am using my second admin.py and urls.py snippets and register django's default model, based on this answer: Django (1.10) override AdminSite

admin.py

from django.contrib.admin import AdminSite
from django.http import HttpResponse
from django.urls import path
from .models import Question
from django.contrib.auth.models import Group, User #add these moduls
from django.contrib.auth.admin import GroupAdmin, UserAdmin #and these

class CustomAdminView(AdminSite):
    def get_urls(self):
        urls = super().get_urls()
        my_urls = [
            path(r'my_view/', self.admin_view(self.my_view))
        ]
        urls = my_urls + urls
        return urls


    def my_view(self, request):
        return HttpResponse("Hello, world.")


custom_admin = CustomAdminView()
custom_admin.register(Question)

#register the default model

custom_admin.register(Group, GroupAdmin)
custom_admin.register(User, UserAdmin)