Classic Secret Santa Problem in 4 lines. Guarantees mathematically that no one will give to themselves and no two will give to each other.

secret-santa.js

function secretSanta(...names) {
  const randomized = [...names].sort(() => Math.random() - Math.random());
  const derrangement = [...randomized]
  derrangement.unshift(derrangement.pop()) // shifting ensures a derrangement of the randomized list
  return Object.fromEntries(randomized.map((name, idx) => [name, derrangement[idx]]))
}

secretSanta('Alice', 'Bob', 'Carter', 'David', 'Emily')
// { Carter: 'Emily', Bob: 'Carter', David: 'Bob', Alice: 'David', Emily: 'Alice' }

For explanation of the math behind how this works: https://www.youtube.com/watch?v=5kC5k5QBqcc

You can also do all this as a one-liner, but it is much less readable:

const secretSanta = 
  (...names) => Object.fromEntries(
    [...names].sort(() => Math.random() - 0.5)
              .map((giver, idx, getters) => [giver, getters[(idx + 1) % getters.length]])
  )

As a party trick, here is that same function minimized (under 100 bytes - it's 94, even room to add a let at the front and still be under):

s=(...n)=>Object.fromEntries(n.sort(_=>Math.random()-.5).map((x,i,a)=>[x,a[(i+1)%a.length]]))

Unrelated

FizzBuzz in 39 bytes:

f=n=>(n%3?'':'Fizz')+(n%5?'':'Buzz')||n

isPalindrome in 46 bytes:

p=s=>[...s].every((c,i)=>c==s[(s.length-i)-1])