When does $\exists x \forall y \phi(x,y) \leftrightarrow \forall y \exists x \phi(x,y)$ hold?

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.

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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: $A\to B \equiv \neg A \lor B$)

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.