Reconstruction of FOL language If we have a FOL language $S$ and we suddenly lost its logical structure $\left(\rightarrow,\wedge,\vee,\bot,\top,\forall x,\exists x\right)$, can we recover it (modulo a logical equivalence) using only the logical consequence relation: $\models$ ? 
Let's consider $S$ the set of formulas of a first order logic and $\bar S$ its quotient set by the logical equivalence relation ($\equiv$). 
All logical connectives over $\bar S$ can be obviously expressed using only $\models$, for example: $$\bar p\rightarrow\bar q=\left\{r\in S\ |\ \forall s\in S,s\models r\Leftrightarrow s,p\models q\right\}$$
Now, and in order to reconstruct quantifiers, let's consider the following set :
$$ \Pi = \left\{\pi\in\big(\bar S^\bar S\big)^*\ \middle\vert\ \begin{array}{l}
    \forall p\in\bar S, (\pi\ p\models p)\wedge(\pi\ \pi\ p=\pi\ p) \\
    \forall p,q\in\bar S, \pi\ (\pi\ p\rightarrow\pi\ q)=\pi\ p\rightarrow\pi\ q\\
    \forall p\in\bar S,\Gamma\subseteq\bar S,\Gamma\models p\Rightarrow \pi\ \Gamma \models \pi\ p
  \end{array}\right\} $$
I think that elements of $\Pi_e=\left\{\pi\in\Pi,\exists!\left(\pi_1,\pi_2\right)\in\Pi^2,\pi=\pi_1\circ\pi_2\right\}$ are universal quantifiers (i.e.
$ p\mapsto\forall x, p[a\leftarrow x] $ where $x$ doesn't occur in $p$ and $a$ is a variable or a constant) plus circular substitutions conjunctions: $$p\mapsto\bigwedge_{i<n}p\left[x_0\leftarrow x_i,...,x_k\leftarrow x_{(k+i \mod n)},...,x_{n-1}\leftarrow x_{(n-1+i \mod n)}\right]$$
Where $(x_i)_i$ can be variables, functions or relations of the same arity.
Is this claim true in intuitionist FOL and in classic FOL ?
If the hole subject is already discussed, could someone please refer me to appropriate articles, and thanks a lot.

EDIT: To be more clear, I'm asking about recovering the logical signature up to a logical equivalence. So we can determine some relation between formulas equivalence classes like: $\overline{\forall x, P(x)}$ is a result of an application of universal quantifier on $\overline{P(x)}$ while $\overline{P(x)\wedge Q(x)}$ is not.

EDIT: And since this is not possible in general, is it possible at least when the non logical signature is finite and contains at least one non-zero arity relation and one non-zero arity function ?
 A: EDIT: Formulas make the situation even easier: any permutation of the variables induces an automorphism of $\mathcal{B}$. For this reason I've left my answer mostly as is, since the interesting situation is when we restrict attention to sentences.
(The above assumes that we don't identify a formula with its universal closure; if we do, then the situation is the same as for sentences since every formula is equivalent to a sentence, namely its universal closure.)

No, you cannot do this in general: when we look at the Boolean algebra $\mathcal{B}$ of semantic equivalence classes of $S$-sentences (which is what you're left with when you forget everything except "$\models$" - incidentally, this is the Lindenbaum algebra of the empty theory over $S$), this algebra has lots of nontrivial automorphisms - these will send sentences to non-equivalent sentences while preserving the $\models$-structure.
Under weak hypotheses on $S$ (let's assume $S$ is countable for now) we can prove this quickly. If $S$ contains two distinct symbols of the same type (e.g. two constant symbols, or two $17$-ary relation symbols, or etc.), then swapping them induces a nontrivial automorphism of $\mathcal{B}$. More complicatedly, if $S$ is infinite then $\mathcal{B}$ is a countable atomless Boolean algebra, and these are homogeneous (given any two elements neither of which is $0$ or $1$, there is an automorphism swapping them); this is proved by a back-and-forth argument, similar to that presented in this proof that any two countable atomless Boolean algebras are isomorphic. Similarly, if $S=\emptyset$ then any permutation of the sentences of the form "there are exactly $n$ elements" induces an automorphism of $\mathcal{B}$.
I believe in fact that we can never perform the reconstruction you want, but I don't have a proof of that yet. Note in particular that we can produces examples of $S$ where some elements of $\mathcal{B}$ are atoms but other elements don't lie above any atoms, so it's neither atomless nor "atomful;" this complicates the picture a bit.

The picture for intuitionistic logic will be similar, except with Heyting algebras taking the role of Boolean algebras. In general, any logic allowing a significant amount of reconstruction along the lines you're proposing would have to be very "algebraically restrictive" (in the sense of algebraic logic), and I'm not aware of any natural such logics.
