(Evolutionary Game Theory) Argue informally there can no longer be an ESS in which only b is played Here is a problem in evolutionary game theory: (So the term I am using should be familiar for people in this field)

The game is called 'Clever Mutants' which is a symmetric two-player game:



For this game, I have managed to find three symmetric Nash Equilibria. 
They are $(a,a), (b,b)$ and $((1/4,3/4),(1/4,3/4))$, in which the first two are evolutionary stable (ES) and the last one is not. 


  Now, suppose that mutants have a 'secret handshake'. That is, suppose
  that mutants can recognize other mutants and play different pure strategies against normal
  and mutant opponents. For example, a mutant could play $b$ against another mutant but
  play $a$ against a non-mutant. Argue informally there can no longer be an $ESS$ (Evolutionary Stable Strategy) in which
  only $b$ is played.


I don't know how to argue for that statement, even informally. Can someone help me, please? Thanks so much.
The "secret handshake" idea was first formalized in economics by Robson (1990).

Suppose initially the equilibrium is $(b,b)$. Since the normal population cannot recognize the mutants, they'd always play $b$. 

When the mutants enter the population, because they can recognize their own kind through a secret handshake, they can play $a$ against one of their own (when there is a secret handshake), and play $b$ against the normal population (no secret handshake). This way, the mutants are getting $3$ each time they meet another mutant and $1$ each time they meet a normal. Therefore, the mutant's strategy is a best response to the normal's strategy, but a better response to itself than the normal's strategy is. As a result, the normal's strategy fails to be evolutionarilly stable.