What's the difference between $\exists v_1(\forall v_1((v_1\cdot v_1)\equiv v_1))$ and $\forall v_1(\exists v_1((v_1\cdot v_1)\equiv v_1))$?
Let $\mathcal {L}$ be the language $\{\mathbf {1},+,\cdot,$exp$,<\}$ where $\mathbf {1}$ is a constant symbol, $+,\cdot$, exp are binary function symbols, and $<$ is a binary relation symbol. Let $\mathcal {N}$ be the $\mathcal {L}$-structure $(\mathbb {N},\mathbf {1}^{\mathcal {N}},+^{\mathcal {N}},\cdot^{\mathcal {N}},$ exp$^{\mathcal {N}},<^{\mathcal {N}})$, where $\mathbf {1}^{\mathcal {N}},+^{\mathcal {N}},\cdot^{\mathcal {N}},<^{\mathcal {N}}$ are the usual operations, and exp$^{\mathcal {N}}(n,m)=n^m$.
My understanding is as follows:
$\exists v_1(\forall v_1((v_1\cdot v_1)\equiv v_1))$ means that $\exists n\in \mathbb {N}$ such that $\forall n((n\cdot n)\equiv n$) is true. However, $\forall n((n\cdot n)\equiv n)$ is clearly not true in $\mathbb {N}$. For example, $2\cdot 2=4\neq 2$. Thus, we conclude that $\mathcal {N}\not\models \phi$.
$\forall v_1(\exists v_1((v_1\cdot v_1)\equiv v_1))$ means that $\forall n\in \mathbb {N}$ such that $\exists n((n\cdot n)\equiv n)$ is true. This is indeed true since $1\cdot 1=1$. Thus, we conclude that $\mathcal {N}\models \phi$.
Correct. Notice that in both sentences, the frst quantifier does not quantify any of the variables in the body; they are so-called null quantifiers, and null quantifiers can be removed without changing the truth-conditions of the sentence.
In other words, $\exists v_1 \forall v_1 (v_1 \cdot v_1=v_1)$ is equivalent to $\forall v_1 (v_1 \cdot v_1 =v_1)$,while $\forall v_1 \exists v_1 (v_1 \cdot v_1 = v_1)$ is equivalent to $\exists v_1 (v_1 \cdot v_1 = v_1)$, and your interpretation would indeed be a good way to show that these are not equivalent.