When I open certain small external community that i am subscribbed to https://lemmings.world/c/[email protected] , i see that moderator list is empty, yet when i visit same community via its parent domain https://lemmy.ml/c/session , i can see one moderator name. Clicking it, cause only page header and footer with no content https://lemmy.ml/u/Seb3thehacker , reloading page (F5) cause error couldnt_find_person. I am not logged in/nor registered on that external instance (lemmy.ml).

So this is confusing, how to become a moderator of an external community? Docs says that i do not need to register on it: “Community moderation can be done over federation, you don’t need to be registered on the same instance where the community is hosted”. Yet it does not explain steps to do. As I have mentioned, i am unable to contact current moderator. I guess one of my options may be to wait and see if current mod find me worthy and adds me as a mod. But i would like to know other options.

  • hetzlemmingsworld@lemmings.worldOP
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    5 months ago

    I guess You mean to create new mod rights request discussion topic inside the community, where i want mod rights (seems like an unsolicited way that pings and spends time of all members)

    Regarding contacting instance admins in case community has no active mods, i assume i go to parent instance (in this case https://lemmy.ml/ ) and scroll down to see the list of “admins:” in the sidebar. I click one, it says “You are not logged in. If you use a Fediverse account that is able to follow users, you can follow this user.” I am unsure how to follow ext. user via my home instance yet i have found this kind of URL: https://lemmings.world/u/[email protected] (assuming my instance lemmings.world) and on it is a New message button that seems to be working. So it does not seems to be easy to contact custom external instance user who’s post i can not see on my instance. UPDATE: I can do it by using search icon and pasting: @[email protected] (for the community, i use [email protected]) - this method is not apparent to a newbie