This can be achieved with the most generic of group by operations.

const arr = ['a', 'b', 'c', 'a', 'b', 'c', 'd'];

const grouped = Object.values(arr.reduce((a, n) => ((a[n] ??= []).push(n), a), {}));

console.log(grouped);

The idea is to sort the array in ascending order, then iterate over it and take if the last char and the current char are the same and put them in an inner-array else create new inner-array of that char, do this process of accumulating till for loop iterate overall characters.

["a","a","b","b","c","c","d"]
//sort and do algorithm 
["a","a","b","b","c","c","d"]
//^---^   ^---^
[["a","a"],["b","b"],["c","c"],["d"]]

Implementation:

const arr = ["a", "b", "c", "a", "b", "c", "d"];
const chars = arr.sort((a, b) => a.localeCompare(b));
console.log(chars);
let res = [[]],
  lastChar = chars[0];
for (char of chars) {
  if (char == lastChar) {
    res[res.length - 1].push(char);
  } else {
    res.push([char]);
    lastChar = char;
  }
}

Result:

console.log(res); //[["a","a"],["b","b"],["c","c"],["d"]]

This is one way to do it. More explicit, but easier to understand and translate to other languages as well. Time: O(n), Space: O(n), n is number of elements in array

function process(arr) {
    const map = arr.reduce((acc, e) => {
        if (!acc.has(e)) {
            acc.set(e, 0);
        }
        acc.set(e, acc.get(e) + 1);
        return acc;
    }, new Map())

    const res = [];
    for (const[k, v] of map.entries()) {
        const localRes = [];
        for (let i = 1; i <= v; i++) {
            localRes.push(k);
        }
        res.push(localRes);
    }
    return res;
}

const arr = ['a', 'b', 'c', 'a', 'b', 'c', 'd']
console.log(process(arr));

Result:

[ [ 'a', 'a' ], [ 'b', 'b' ], [ 'c', 'c' ], [ 'd' ] ]