No, you are not right. The documentation on readonly struct is perfectly correct. Your question contains a broken link, the correct URL of this documentation page and section is this.

The explanation is pretty simple. Consider this example:

    readonly struct A {
        internal A(int first, string second) { First = first; Second = second; }
        internal readonly int First { get; init; }
        internal readonly string Second { get; init; }
    }

    static class ReadonlyUsage {
        static A value = new(0, "original value");
        internal static void Demo() {
            A a = new(1, "constant value");
            //a.First = 12; // will fail to compile
            a = new(2, "another value");
            // this is mutable:
            value = new(3, "one more value");
            //value.First = 13; // will fail to compile
            // still mutable:
            value = new(4, "How much longer?!");
        }
    }

What is mutable here? Nothing except ReadonlyUsage.value, and it makes ReadonlyUsage also mutable. It has nothing to do with struct A — it is immutable. Exactly as System.String.

Let's see: what about the assignment to A a? Does this assignment make ReadonlyUsage mutable? No. You assign a new value (and struct A is a value type) to a stack variable. It exists only in the stack and will be removed after the exit of a current stack frame. The stack does not belong to the class, it belongs to the thread calling ReadonlyUsage.Demo. The assignment does not modify anything in the Demo class if it was an instance class. We simply create a new struct A instance and never modify any of the already existing instances, that's why it is called immutable.

Now, what about the assignment to ReadonlyUsage.value? It is possible, because ReadonlyUsage is mutable. Again, we create new instances of struct A and never modify any of the already existing instances.

How to prevent the assignment to ReadonlyUsage.value? Easy: make it readonly, too:

static readonly A value = new(0, "original value");

That's it. I hope the misunderstanding is dismissed. Will you accept the answer then? If not, please ask further questions.