You can use just regular sort

const list = [{key: "Name",textColor: "red"},{key: "Age",textColor: "green"},{key: "Occupation",textColor: "yellow"}];
const newOrder = ["Occupation", "Name", "Age"];

const result = list.sort(({key: a}, {key: b}) => newOrder.indexOf(a) - newOrder.indexOf(b));

console.log(result);

.as-console-wrapper{min-height: 100%!important; top: 0}

Your list can be reformatted to a regular javascript object in which key is the property name, and textColor is the value:

const toObject = kvps => Object.fromEntries(kvps.map(kvp => [ kvp.key, kvp.textColor ]));

With a given array of keys, you can pick values from that object like so:

const fromObject = (order, obj) => order.map(key => ({ key, textColor: obj[key] }));

Chain the two together, and you can reorder any list of key value pairs:

const list = [{
    key: "Name",
    textColor: "red"
  },{
    key: "Age",
    textColor: "green"
  },{
    key: "Occupation",
    textColor: "yellow"
}]


const toObject = kvps => Object.fromEntries(kvps.map(kvp => [ kvp.key, kvp.textColor ]));
const fromObject = (order, obj) => order.map(key => ({ key, textColor: obj[key] }));
const reorder = (order, kvps) => fromObject(order, toObject(kvps));


const newList = reorder(["Occupation", "Name", "Age"], list);

console.log(
  newList
)

_{Edit: if the sizes of your list and order arrays are small, you probably want to go with the much easier to read approach suggested by Jon Webb in one of the other answers. 🙂 I tried to keep my solution to an O(n + m) complexity rather than O(n * m), (n = list size, m = order size) but it's probably not worth the added complexity.}

You can iterate on the "order" list, finding the corresponding item in the original list, and push the item to the new list in that order:

const orderList = (list, order) => {
  const newList = [];

  for (key of order) {
    const item = list.find((obj) => obj.key == key);
    if (item) newList.push(item);
  }

  return newList;
}