displayName已过时，但placeId尚未过时。
下面是一个突出显示意大利的示例。要获得地点ID，可以使用Place ID finder。我添加了一个对Geocoder的调用，以便能够使Map适合要素。

let map;
let featureLayer;

function initMap() {

  let geocoder = new google.maps.Geocoder();
  map = new google.maps.Map(document.getElementById("map"), {
    center: {
      lat: 0,
      lng: 0
    },
    zoom: 13,
    mapId: "8ec1ff859561face", // You need to create your own map ID
  });

  featureLayer = map.getFeatureLayer("COUNTRY");

  // Define the styling options
  const featureStyleOptions = {
    strokeColor: "#00FF00",
    strokeOpacity: 1,
    strokeWeight: 1,
    fillColor: '#519400',
    fillOpacity: 1,
  };

  // Below Place ID is "Italy"
  let placeId = 'ChIJA9KNRIL-1BIRb15jJFz1LOI';

  geocoder.geocode({
      placeId: placeId
    })
    .then(({
      results
    }) => {
      map.fitBounds(results[0].geometry.viewport);
    })
    .catch((e) => {
      console.log('Geocoder failed due to: ' + e)
    });

  // Apply the style to a single boundary.
  featureLayer.style = (options) => {
    if (options.feature.placeId == placeId) {
      return featureStyleOptions;
    }
  };
}

#map {
  height: 180px;
}

<div id="map"></div>
<script src="https://maps.googleapis.com/maps/api/js?key=AIzaSyCkUOdZ5y7hMm0yrcCQoCvLwzdM6M8s5qk&callback=initMap&v=beta" defer></script>

关于fetchPlace()，文档规定：
不要从FeatureStyleFunction调用它，因为只支持同步FeatureStyleFunctions。