Suppose $\Gamma_1(V_1, E_1)$ and $\Gamma_2(V_2, E_2)$ are simple graphs with countably many vertices. And suppose $A_1$ and $A_2$ are initially empty. Suppose two players play the following game: each turn, the first player choses either to add a vertex from $V_1$ to $A_1$ or a vertex from $V_2$ to $A_2$. Then the second player also choses either to add a vertex from $V_1$ to $A_1$ or a vertex from $V_2$ to $A_2$. After it, if the subgraphs induced by $A_1$ and $A_2$ are not isomorphic, the first player wins. Otherwise the game continues. The second player wins, if they are able to keep the game going on indefinitely.
Is it true, that the second player has a winning strategy iff $\Gamma_1 \cong \Gamma_2$?
I know, that if $\Gamma_1 \cong \Gamma_2$ then the second player can prolong the game indefinitely by simply copying the moves of the first player on the different graph. But what’s about the converse? Is it true?