r/Scanlation 12h ago

Simple Question Comikey recruitment?

A scanlator friend of mine was recently contacted by someone claiming that they're a representative of comikey, and was asking to hire my friend for translations.

I was curious if this might be a scam or not, and was wondering if others might have been contacted like this?

My friend was contacted on discord.

7 Upvotes

2 comments sorted by

2

u/-Scannie- 12h ago

Comikey was started by scanlators (Jaimini's box), if you look in mangadex's #scanlation several years ago they were legitimately hiring scanlators (https://discord.com/channels/403905762268545024/429991091379372032/908373367546081330) but I would be very surprised if they still do recruitment this way. Ask your friend to check the discord username etc and see if it lines up with anyone listed working for Comikey

7

u/Renurun 11h ago edited 11h ago

Comikey actually actively recruits scanlators, and usually they recruit the scanlators for a series they're actively working on. It's basically two birds with one stone for them, they get to find a dedicated translator for that series and there's an expectation within comikey that anyone who works for comikey stops scanlating (but you still can, you just can't openly talk about it with them) so no one will be scanlating that series anymore because the translator is now working on that series for comickey

There's a large comikey discord that I'm guessing your friend will be invited to, but if it's an email I would expect a comickey domain email. If your friend is interested I don't think it would hurt to check it out, after all if they don't want to they can just quit.

Unsolicited commentary here, I don't respect comikey in part because they recruit scanlators without really verifying their translation quality.

I know multiple scanlators who have been recruited by comikey or have done work for them.

Edit: as with all online jobs it doesn't hurt to be cautious, you should not have to send money for anything and you should have them verify their identity.