Did you reply to the wrong comment, I'm not sure it follows from what I posted?

You can't force a collision for the function f(x) = x, by definition it has no collisions. It's not just collision-resistant, it's actually collision proof!

hash(x) = x is of course not a cryptographic hash, but it also has the advantage that it's unlikely to be confused for one if anyone looks at the output.

Nitpick: it’s still going to collide when taking the modulo to decide what bucket to put it in. If you know that the hashmap has 1000 buckets, and you spam it with multiples of 1000, you could make it put every key into the same bucket and have O(1) lookups turn into O(n). That doesn’t happen with real hash functions with random seeds.

It's not about collisions of the hash function itself.

Every hashtable implementation will put the hash value through some sort of modulo because you generally don't want to waste memory storing the key 5726591 at index 5726591 in an array.

So if you know how the implementation works and the hash function is predictable you can keep providing the program with values that will consistently go into the same bucket resulting in linear lookups and insertions.