Determinacy and Ehrenfeucht-Fraïssé Let us consider $L$ a first order language with no function symbols, and $\mathfrak{A,B}$ two $L$-structures.
We consider this variant of the Ehrenfeucht-Fraïssé game : player I (Spoiler) picks at each step an element of $A$ or $B$ (the underlying sets of the given structure), and when it's his turn, player II (Duplicator) does the same.
We add the following rules, that can be modelled by a tree : 
If I picks an element of $A$ at his turn, then II must pick an element of $B$ and similarly if I picks an element of $B$, II must pick an element of $A$. 
Now II wins if after each of his turns, there is an isomorphism between the induced structures of the elements chosen from $A$ and those chosen from $B$.
Notice that thus condition can be expressed as "the sequence obtained belongs to the body of the tree $T$" where $T$ is a subtree of the tree representing the rules, and thus this game is a Gale-Stewart game with rules where the payoff set is closed, and therefore by a theorem of Gale and Stewart, it is determined (assuming the axiom of choice).
One can see that if Duplicator has a winning strategy in all Ehrenfeucht-Fraïssé games $G_n(\mathfrak{A,B})$ then he has one in this variant, and conversely. In such a case, it is known that $\mathfrak{A}$ and $\mathfrak{B}$ are elementarily equivalent.
Moreover, if the set of relation symbols is finite, and if $\mathfrak{A,B}$ are elementarily equivalent, then Duplicator has a winning strategy in all games $G_n(\mathfrak{A,B})$ and therefore he has one in the variant.
Therefore we have : if the set of relation symbols is finite, then Duplicator wins (in the variant) if and only if $\mathfrak{A}$ and $\mathfrak{B}$ are elementarily equivalent, and therefore (the game being determined) Spoiler wins if and only if they are not.
Now my questions are : 
Q1: Is this variant interesting beyond simply "It's only one game, as opposed to an infinity of games" ? If yes, has it been studied before ? Does the Gale-Stewart theorem (and other theorems concerning Gale-Stewart games) bring any new information to Ehrenfeucht-Fraïssé games ?
Q2 : What happens when there's an infinite set of relation symbols ? 
EDIT : If it wasn't clear, in my variant, the isomorphisms don't have to be extensions of each other, they can change at each step. 
However I now realize that I went too fast with this sentence "and conversely" (when I mentioned the strategies for $G_n(\mathfrak{A,B})$, and it may actually be false. It would be true if Duplicator played the same way on the first $n$ rounds of $G_n(\mathfrak{A,B})$ and $G_m(\mathfrak{A,B})$ for $n<m$ (given a play by Spoiler) but maybe the strategy of Duplicator can change according to $n$
 A: Let's call your game $G_\omega(\mathfrak{A},\mathfrak{B})$. It's clear that a winning strategy for Duplicator in $G_\omega(\mathfrak{A},\mathfrak{B})$ provides a winning strategy in $G_n(\mathfrak{A},\mathfrak{B})$ for all $n$. But the converse is not true - essentially, this is because the winning strategies for the games of length $n$ for various $n$ might not agree, and therefore might not cohere to a strategy in the infinite game.
For an explicit counterexample, let $\mathfrak{A} = (\mathbb{N},\leq)$ and let $\mathfrak{B} = (\mathbb{N}+\mathbb{Z},\leq)$. The theory of a discrete linear order with a least element is complete, so these structures are elementarily equivalent. But Sploiler has a winning strategy in $G_\omega(\mathfrak{A},\mathfrak{B})$: First, Spoiler chooses any element in the copy of $\mathbb{Z}$ in $\mathfrak{B}$. Duplicator is forced to respond with an element $k$ in $\mathfrak{A}$. Then, in the next $k$ turns, Spoiler chooses the elements $0,1,\dots,k-1$ in the copy of $\mathbb{N}$ in $\mathfrak{B}$, and Duplicator is forced to respond with $0,1,\dots,k-1$ in $\mathfrak{A}$. When Spoiler chooses $k$ in the next turn, Duplicator has no way to extend the partial isomorphism.
In fact, Duplicator has a winning strategy in $G_\omega(\mathfrak{A},\mathfrak{B})$ if and only if $\mathfrak{A}\equiv_{\infty,\omega}\mathfrak{B}$, meaning that $\mathfrak{A}$ and $\mathfrak{B}$ agree on every sentence of the infinitary logic $\mathcal{L}_{\infty,\omega}$. Needless to say, this is a much stronger condition than elementary equivalence. This is essentially Karp's Theorem. See Hodges Model Theory Section 3.5 (actually, the whole of Chapter 3 on EF games is quite informative).
