When does $\exists x \forall y \phi(x,y) \leftrightarrow \forall y \exists x \phi(x,y)$ hold? Clearly, $\exists x \forall y \phi(x,y) \to \forall y \exists x \phi(x,y)$ is a tautology. However, the other way around is not a tautology: $\forall y \exists x \phi(x,y) \to \exists x \forall y \phi(x,y)$.
Nevertheless, I am interested in the case that this holds. That is, I am interested in the set of structures and formulae $\phi(x,y)$ where the following holds:
$$\exists x \forall y \phi(x,y) \leftrightarrow \forall y \exists x \phi(x,y)$$
Does there exist an analysis of structures where this statement holds? Are there interesting things to be said about them? 
EDIT: Note that this does not require the statement to hold for arbitrary $\phi(x,y)$. i.e. if we need to restrict $\phi(x,y)$ in some way to get something interesting, then I still like to know about it.
 A: I don't know if this answers your questions, but it's too long for a comment.
Put all of your of your statements $\phi(x,y)$ in a 2D array (I assumed countable statements, just for convenience):
\begin{array}{c c c c c}
\phi(x_0,y_0) & \phi(x_1,y_0) & \phi(x_2,y_0) & \phi(x_3,y_0) & \dots\\
\phi(x_0,y_1) & \phi(x_1,y_1) & \phi(x_2,y_1) & \phi(x_3,y_1) & \dots \\
\phi(x_0,y_2) & \phi(x_1,y_2) & \phi(x_2,y_2) & \phi(x_3,y_2) & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
then you'll notice that the first statement $\exists x\forall y$ is saying that a column in this array is correct. In the second case $\forall y \exists x$ you are simply saying that each row has at least one correct statement.
Thus the two are equal either if you only have $1$ column, or if there is at least one column where each row has a correct statement. 
A: Hint 1: It holds when the domain of discourse is a single object. 
Hint 2:  It does not hold when the domain of discourse is empty.

EDIT: $\forall x \exists y \phi(x,y) \to \exists y \forall x \phi(x,y)$ will hold if and only if: 

$\exists x \forall y \neg \phi (x,y)$ (making the antecedent false)

OR 

$\exists y \forall x \phi(x,y)$ (making the consequent true).

(17 line proof using: $A\to B \equiv \neg A \lor B$)
A: It holds (also) in the case $\phi(x,y)$ is of the form $p(x)\vee q(y)$ or $p(x)\wedge q(y)$. See Prenex Normal Forms. For example
\begin{align}
&\exists x\ [\ \forall y\ (\ p(x)\vee q(y)\ )\ ] \\
\iff & \exists x\ [\ p(x)\vee (\ \forall y\ q(y)\ )\ ] \\
\iff & (\ \exists x\  p(x)\ )\vee (\ \forall y\ q(y)\ ) \\
\iff & \forall y\ [\ (\ \exists x\  p(x)\ )\vee  q(y)\ ] \\
\iff & \forall y\ [\ \exists x\  (\ p(x)\vee  q(y)\ )\ ] \\
\end{align}
