First, deserialize the JSON into the object and put all the objects into a list and then serialize the list object to JSON.

The actual way is using JS Object.assign.

Object.assign(target, ...sources)

MDN Link

There is another object spread operator which is proposed for ES7 and can be used with Babel plugins.

 Obj = {...sourceObj1, ...sourceObj2}

If you really want to reduce the number of HTTP requests from the client, one possible way is like this:

1. Build a server side proxy that get JSon files from the third party data sources
2. Merge your 3 JSon files in one single JSon file
3. Make a single $.getJSON() call to your single JSon from client

To merge Json files, look at this post.

Upon first appearance, the word "merg" leads one to think you need to use .extend, which is the proper jQuery way to "merge" JSON objects. However, $.extend(true, {}, json1, json2); will cause all values sharing the same key name to be overridden by the latest supplied in the params. As review of your question shows, this is undesired.

What you seek is a simple javascript function known as .concat. Which would work like:

var finalObj = json1.concat(json2);

While this is not a native jQuery function, you could easily add it to the jQuery library for simple future use as follows:

;(function($) {
    if (!$.concat) {
        $.extend({
            concat: function() {
                return Array.prototype.concat.apply([], arguments);
            }
        });
    }
})(jQuery);

And then recall it as desired like:

var finalObj = $.concat(json1, json2);

You can also use it for multiple array objects of this type with a like:

var finalObj = $.concat(json1, json2, json3, json4, json5, ....);

And if you really want it jQuery style and very short and sweet (aka minified)

;(function(a){a.concat||a.extend({concat:function(){return Array.prototype.concat.apply([],arguments);}})})(jQuery);

;(function($){$.concat||$.extend({concat:function(){return Array.prototype.concat.apply([],arguments);}})})(jQuery);

$(function() {
    var json1 = [{id:1, name: 'xxx'}],
        json2 = [{id:2, name: 'xyz'}],
        json3 = [{id:3, name: 'xyy'}],
        json4 = [{id:4, name: 'xzy'}],
        json5 = [{id:5, name: 'zxy'}];
    
    console.log(Array(10).join('-')+'(json1, json2, json3)'+Array(10).join('-'));
    console.log($.concat(json1, json2, json3));
    console.log(Array(10).join('-')+'(json1, json2, json3, json4, json5)'+Array(10).join('-'));
    console.log($.concat(json1, json2, json3, json4, json5));
    console.log(Array(10).join('-')+'(json4, json1, json2, json5)'+Array(10).join('-'));
    console.log($.concat(json4, json1, json2, json5));
});

center { padding: 3em; }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<center>See Console Log</center>

jsFiddle