how to group in mongoDB and return all fields in result

When you group data on any database, it means you want to perform accumulated operation on the required field and the other field which will not be include in accumulated operation will be used in group like

 db.collection.aggregate([{
 $group: {
   _id: { field1: "", field1: "" },
   acc: { $sum: 1 }
 }}]

here in _id object will contains all other fields which you want to hold.

for your data you can try this

db.collection.aggregate([{
    $group: {
        _id: "$name",
        rating: { $first: "$rating" },
        tags: { $first: "$tag" },
        docid: { $first: "$_id" }
    }
},
{
    $project: {
        _id: "$docid",
        name: "$_id",
        rating: 1,
        tags: 1
    }
}])

You can use below aggregation query.

$$ROOT to keep the whole document per each name followed by $replaceRoot to promote the document to the top.

db.col.aggregate([
  {"$group":{"_id":"$name","doc":{"$first":"$$ROOT"}}},
  {"$replaceRoot":{"newRoot":"$doc"}}
])

You can use this query

db.col.aggregate([
                        {"$group" : {"_id" : "$name","data" : {"$first" : "$$ROOT"}}},
                        {"$project" : {
                            "tags" : "$data.tags",
                            "name" : "$data.name",
                            "rating" : "$data.rating",
                            "_id" : "$data._id"
                            }
                        }])

user2683814's solution worked for me but in my case, I have a counter accumulator when we replace the newRoot object, the count field is missing in the final stage so I've used $mergeObjects operator to get my count field back.

db.collection.aggregate([
 {
  $group: {
    _id: '$product',
    detail: { $first: '$$ROOT' },
    count: {
      $sum: 1,
    },
  },
},
{
  $replaceRoot: {
    newRoot: { $mergeObjects: [{ count: '$count' }, '$detail'] },
  },
}])