Which answer did this user write? This is an answer-chaining challenge where each answer must take an integer \$N\$ as input and return the 1-indexed ID of the answer which was written by the user whose ID is \$N\$.
You may want to sort by oldest to get the answers in the logical order.
Rules

I will post the first answer.
You are only allowed to post one answer, so that a user ID is linked to one and only one answer ID.
You are only supposed to support user IDs that were used in the previous answers and your own user ID. (So your code may return anything or crash when given an unknown user ID.)
The maximum code size is set to 50 bytes. So the challenge will end when nobody is able to add a new ID without exceeding this size.
You may use any language, no matter which ones were used before. It is recommended to use verbose languages early in the chain, while they still have a chance to compete.
The last valid answer in the chain on January 19, 2022 at 18:00 UTC wins the challenge.

Please include in your post the list of ID pairs, updated with your new entry:
[your_user_ID] -&gt; [your_answer_ID]

Also make sure to include the ID of your answer in the header.
Examples
The first answer is written by user Arnauld (58563):
1. JavaScript (ES6), 4 bytes
_=&gt;1

Try it online!
Updated list:
58563 -&gt; 1

This code returns 1 when given 58563 (my user ID) ... or any other input. That's fine because 1 is the only possible output at this point (see rule #3).
The second answer is written by user JohnDoe (123456) (who actually doesn't exist yet):
2. Python 2, 14 bytes
lambda n:2-n%2

Try it online!
Updated list:
58563 -&gt; 1
123456 -&gt; 2

This code returns 1 when given 58563 or 2 when given 123456.
7. Julia, 17 bytes
works due to Int64 overflowing to negative numbers

x-&gt;(x-8479)^6%5+5

Try it online!
Updated list:
58563 -&gt; 1
80214 -&gt; 2
95792 -&gt; 3
89930 -&gt; 4
86147 -&gt; 5
97857 -&gt; 6
98541 -&gt; 7

I found those numbers by guessing and brute-forcing :

for i in Iterators.product(-9999:9999,1:20,1:20)
    d = diff(((L.+i[1]) .^ i[3]) .% i[2])
    if all(==(-1), d) || all(==(1),d)
        println(i)
        println(((L.+i[1]) .^ i[3]) .% i[2])
        break
    end
end

Try it online!