I think this example illustrates your issue:

    class FooAction(argparse.Action):
        def __call__(self, parser, namespace, values, 
            option_string = None):
            if values=='Q':
                setattr(namespace, self.dest, 
                    {'type':values, 'quick':True, 'slow':False})
            else:
                setattr(namespace, self.dest, 
                    {'type':values, 'quick':False, 'slow':True})

    def bar(string):
        return string.upper()[0]

    parser = argparse.ArgumentParser()
    parser.add_argument('-a','--algorithm', choices=['Q','S'],
        default='s', action=FooAction, type=bar)
    parser.add_argument('foo', choices=['Q','S'],
        default='s', action=FooAction, nargs='?', type=bar)
    print(parser.parse_args([]))
    # Namespace(algorithm='S',
    #           foo={'slow': True, 'type': 'S', 'quick': False})

For both the optional and positional, the lower case default is converted into an upper case letter.

For the positional with nargs='?', the action is called with the converted default value.

But for the optional, the converted default is assigned to the dest
   algorithm='S'
but action is not called, which would have assigned
   algorithm={'type': 'S', 'quick': False, 'slow': True}

That is consistent with what I remember from the code.  A next function 'take_action' is called only if there is an appropriate argument string.  For this optional that would be '-a' (or the likes).  For the '?' positional and empty argument list enough.

There is a relatively recent change that delayed when a default was converted by type (so a filetype wouldn't be opened unnecessarily).  But I can't think anyway that the action would be invoked on the default in the absence of '-a'.

In this example, would it be possible shift code from the custom action to the custom type?