With Beyond Compare you can use in the File Formats-Settings the XML Sort Conversion. With this option the XML children will be sorted before the diff.

A trial / portable version of Beyond Compare is available.

My original answer is outdated. If I would have to build it again i would use xmlunit 2 and xmlunit-matchers. Please note that for xml unit a different order is always 'similar' not equals.

@Test
public void testXmlUnit() {
    String myControlXML = "<test><elem>a</elem><elem>b</elem></test>";
    String expected = "<test><elem>b</elem><elem>a</elem></test>";
    assertThat(myControlXML, isSimilarTo(expected)
            .withNodeMatcher(new DefaultNodeMatcher(ElementSelectors.byNameAndText)));
    //In case you wan't to ignore whitespaces add ignoreWhitespace().normalizeWhitespace()
    assertThat(myControlXML, isSimilarTo(expected)
        .ignoreWhitespace()
        .normalizeWhitespace()
        .withNodeMatcher(new DefaultNodeMatcher(ElementSelectors.byNameAndText)));
}  

If somebody still want't to use a pure java implementation here it is. This implementation extracts the content from xml and compares the list ignoring order.

public static Document loadXMLFromString(String xml) throws Exception {
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder = factory.newDocumentBuilder();
    InputSource is = new InputSource(new StringReader(xml));
    return builder.parse(is);
}

@Test
public void test() throws Exception {
    Document doc = loadXMLFromString("<test>\n" +
            "  <elem>b</elem>\n" +
            "  <elem>a</elem>\n" +
            "</test>");
    XPathFactory xPathfactory = XPathFactory.newInstance();
    XPath xpath = xPathfactory.newXPath();
    XPathExpression expr = xpath.compile("//test//elem");
    NodeList all = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
    List<String> values = new ArrayList<>();
    if (all != null && all.getLength() > 0) {
        for (int i = 0; i < all.getLength(); i++) {
            values.add(all.item(i).getTextContent());
        }
    }
    Set<String> expected = new HashSet<>(Arrays.asList("a", "b"));
    assertThat("List equality without order",
            values, containsInAnyOrder(expected.toArray()));
}

You might need to add something along the lines of

.withDifferenceEvaluator(((comparison, outcome) -> {
    if (outcome == ComparisonResult.DIFFERENT && 
        comparison.getType() == ComparisonType.CHILD_NODELIST_SEQUENCE) {
           return ComparisonResult.EQUAL;
    }
return outcome;
})).build();

to your Diff builder

for me the solution mentioned above will not work as compareNodeLists has this hardcoded in DOMDifferenceEngine.compareNodes()

new Comparison(ComparisonType.CHILD_NODELIST_SEQUENCE...

i have raised a new ticket for this though bear in mind it could just be my lack of understading :-)

https://github.com/xmlunit/xmlunit/issues/258

Good answer here:

Question: How can I diff two XML files? | Super User

Answer: How can I diff two XML files? | Super User

$ xmllint --format --exc-c14n one.xml > 1.xml
$ xmllint --format --exc-c14n two.xml > 2.xml
$ diff 1.xml 2.xml

Apologies for any failure to adhere to serverfault conventions ... I'm sure someone will let me know and I will amend appropriately.

You're requesting a sort based on the sequence of attributes in the elements being sorted. But your top-level tag elements here have only one attribute: name. If you want multiple tag elements with name="BBB" to sort differently, you need to give them distinct sort keys.

In your example, I'd try something like select="concat(name(), @name, name(*[1]), *[1]/@name)" -- but this is a very shallow key. It uses values from the first child in the input, but the children may shift position during the process. You may be able (knowing your data better than I do) to calculate a good key for each element in a single pass, or you may just need several passes.