Showing equivalence of set of solutions of two linear programming Setup of the problem (updated thanks to comments below)
Consider six finite sets of real numbers each with cardinality $4$
$$
\mathcal{A}\equiv \{a_1,a_2,a_3,a_4\}\text{, }\text{ }\tilde{\mathcal{A}}\equiv \{\tilde{a}_1,\tilde{a}_2,\tilde{a}_3,\tilde{a}_4\}
$$
$$
\mathcal{B}\equiv \{b_1,b_2,b_3,b_4\}\text{, }\text{ }\tilde{\mathcal{B}}\equiv \{\tilde{b}_1,\tilde{b}_2,\tilde{b}_3,\tilde{b}_4\}
$$
$$
\mathcal{C}\equiv \{c_1,c_2,c_3,c_4\}\text{, }\text{ }\tilde{\mathcal{C}}\equiv \{\tilde{c}_1,\tilde{c}_2,\tilde{c}_3,\tilde{c}_4\}
$$
Let $$\mathcal{G}\equiv \mathcal{A}\times \mathcal{B}\times \mathcal{C}\text{, }\text{ }\tilde{\mathcal{G}}\equiv \tilde{\mathcal{A}}\times \tilde{\mathcal{B}}\times \tilde{\mathcal{C}}$$ where $\times$ denotes Cartesian product. Hence $\mathcal{G}$ is a collection of $4^3$ $3$-tuples, i.e., $\mathcal{G}\equiv \{(a_1,b_1,c_1), (a_2, b_1, c_1),...\}$ 
Consider the 4-tuples (i.e., the order matters)
$$
A\equiv (a_1,a_2,a_3,a_4)\text{, }\text{ }\tilde{A}\equiv (\tilde{a}_1,\tilde{a}_2,\tilde{a}_3,\tilde{a}_4)
$$
$$
B\equiv (b_1,b_2,b_3,b_4)\text{, }\text{ }\tilde{B}\equiv (\tilde{b}_1,\tilde{b}_2,\tilde{b}_3,\tilde{b}_4)
$$
$$
C\equiv (c_1,c_2,c_3,c_4)\text{, }\text{ }\tilde{C}\equiv (\tilde{c}_1,\tilde{c}_2,\tilde{c}_3,\tilde{c}_4)
$$
Let $\pi$ be an operator that when applied to a tuple gives: 
(1) the position in the original tuple of each element of the tuple when ordered from smallest to largest
(2) the relation operators ($<$, $=$) 
For example, if $A\equiv (100,99,102,0)$ then $\pi(A)=\{(4,2,1,3), (<,<,<)\}$. If $A\equiv (5,5, \infty, -\infty)$ then $\pi(A)=\{(4,1,2,3), (<,=,<)\}$.
Consider the following 2 linear programmings, where $p_1, p_2, p_3$ are known parameters in $(0,1)$
LP1
\begin{equation}
\begin{aligned}
& \text{Find $g:\mathcal{G}\rightarrow \mathbb{R}$ such that }\\
& (1) \text{ }p_1=1+g(a_1, b_2, c_1)-g(a_2, b_1, c_3)-g(a_4, b_4, c_4) \\
& (2) \text{ }p_2=g(a_3, b_4, c_3)-g(a_4, b_1, c_3)\\
& (3) \text{ }p_3=g(a_1, b_2, c_4)\\
& (4) \text{ } -g(t,q,r)
+g(t', q, r)
+g(t, q', r)
-g(t', q', r)\\
&\hspace{1.5cm}+
g(t, q, r')
-g(t', q, r')
-g(t, q', r')
+g(t', q', r' )\geq 0\\
 &\hspace{5cm} { \text{$\forall \{(t,q,r), (t',q',r')\}\subseteq \mathcal{G}$ s.t. $(t,q,r)\leq (t',q',r')$}}
\end{aligned}
\end{equation}
and 
LP2
\begin{equation}
\begin{aligned}
& \text{Find $g:\tilde{\mathcal{G}}\rightarrow \mathbb{R}$ such that }\\
& (1) \text{ }p_1=1+g(\tilde{a}_1, \tilde{b}_2, \tilde{c}_1)-g(\tilde{a}_2, \tilde{b}_1, \tilde{c}_3)-g(\tilde{a}_4, \tilde{b}_4, \tilde{c}_4) \\
& (2) \text{ }p_2=g(\tilde{a}_3, \tilde{b}_4, \tilde{c}_3)-g(\tilde{a}_4, \tilde{b}_1, \tilde{c}_3)\\
& (3) \text{ }p_3=g(\tilde{a}_1, \tilde{b}_2, \tilde{c}_4)\\
& (4) \text{ }-g(t,q,r)
+g(t', q, r)
+g(t, q', r)
-g(t', q', r)\\
&\hspace{1.5cm}+
g(t, q, r')
-g(t', q, r')
-g(t, q', r')
+g(t', q', r' )\geq 0\\
 &\hspace{5cm} { \text{$\forall \{(t,q,r), (t',q',r')\}\subseteq \tilde{\mathcal{G}}$ s.t. $(t,q,r)\leq (t',q',r')$}}
\end{aligned}
\end{equation}
Let $q_1$ if the set of solutions of LP1 is not empty and zero otherwise.
Let $q_2$ if the set of solutions of LP1 is not empty and zero otherwise.

I would like your help to show that 
Claim: If $\mathcal{G}$ and $\tilde{\mathcal{G}}$ are such that 
$$
\pi(A)=\pi( \tilde{A})\\
\pi(B)=\pi( \tilde{B})\\\\
\pi(C)=\pi( \tilde{C})\\
$$
then $q_1=q_2$.
 A: Firstly, there won't be 16 3-tuples in $\mathcal{G}$, there'll be $4^3=64$.
I'm still not 100% convinced this is the thing you are looking for. You state the problem is to find a function $g$ that satisfies the constraints (that's not really a linear program in the usual sense). If you set $g=0$, it will always be a solution. So the solution set will always be non-empty to both problems, and so $q_1=q_2=0$ always.
If you are actually fixing the function $g$ in advance and then determining the solution set of pair of 3-tuples from $\mathcal{G}$ which satisfy the constraints, then the statement is again always true, regardless of the underlying sets $\mathcal{A},\tilde{\mathcal{A}},\cdots$ etc. because $\{(a_1,b_1,c_1),(a_1,b_1,c_1)\}$ will always be a solution, since the terms with $g$ cancel out to zero.

(16 Jan 2019) based on older statement of the problem
This isn't true, at least in the current form you've presented it.
Consider $\mathcal{A}=\tilde{\mathcal{A}}=\mathcal{B}=\tilde{\mathcal{B}}=\mathcal{C}=(1,2,3,4)$, and $\tilde{\mathcal{C}}=(2,3,4,5)$. Then define $g(t,q,r)=0$. Clearly $\pi(\mathcal{A})=\pi(\tilde{\mathcal{A}})=\pi(\mathcal{B})=\pi(\tilde{\mathcal{B}})=\pi(\mathcal{C})=\pi(\tilde{\mathcal{C}})$. Yet for $\mathsf{LP}1$, $\{(1,1,1),(1,1,1)\}$ is a solution, $\{(1,1,5),(1,1,5)\}$ is not, whereas for $\mathsf{LP}2$, $\{(1,1,1),(1,1,1)\}$ is not a solution, but $\{(1,1,5),(1,1,5)\}$ is. So the solution sets cannot be the same.
