This question has been edited thanks to the useful comments below (apologies for my previous mistakes/confusions):
Consider a function $\Phi: \mathcal{X}\times \mathcal{Y}\subseteq \mathbb{R}^2\rightarrow \mathbb{R}$, where $\mathcal{X}$ and $\mathcal{Y}$ are finite sets and $\{0,1\}\subset \mathcal{X}$, $\{0,1\}\subset \mathcal{Y}$.
Assume
(a) $\Phi(0,y)$ and $\Phi(x,0)=0$ $\forall y\in \mathcal{Y}$ and $\forall x \in \mathcal{X}$
(b) $\Phi(1,1)=1$
Consider a function $\tilde{\Phi}: \mathcal{X}\times \mathcal{Y}\rightarrow \mathbb{R}$. Suppose
(c) $\tilde{\Phi}(0,y)$ and $\tilde{\Phi}(x,0)=0$ $\forall y\in \mathcal{Y}$ and $\forall x \in \mathcal{X}$
(d) $\tilde{\Phi}(1,1)=1$
(I) Does there exist $\tilde{\Phi}$ such that $\Phi$ can be expressed as a strictly increasing transformation of $\tilde{\Phi}$?
(II) If the answer to (I) is YES, which restrictions (alternative to a,b,c,d or in addition to a,b,c,d) would guarantee that there does not exists $\tilde{\Phi}$ such that $\Phi$ be expressed as a strictly increasing transformation of $\tilde{\Phi}$?
[For question (II) I'm looking for a sort of "minimal" set of restrictions. For example, I guess that $\tilde{\Phi}(x,y)=\Phi(x,y)$ $\forall (x,y)\in \mathcal{X}\times \mathcal{Y}$ would make the job, but I would like something weaker]