Two Hausdorff Compactifications are equivalent iff the sets of bounded continuous real valued functions in X which extend to Y and Z coincide. Let $(g, Y)$ and $(h, Z)$ be two Hausdorff compactifications of $X$.
We have that $(g, Y)$ is equivalent to $(h, Z)$ if $\exists \text{ homeomorphism } \pi : Z \to Y$ such that $\pi \circ h = g$ by definition. 
If this is the case, then if $f$ extends to $Y$ so there is a continuous $F:Y \to \Bbb{R}$ such that $F \circ g = f$ we see that $F \circ \pi$ extends $f$ to $Z$. Similarly the symmetrical case, so we have that if $(h, Z)$ and $(g, Y)$ are equivalent then the sets of bounded continuous real valued functions on $X$ which extend $Y$ and $Z$ coincide. 
I am really struggling to show the converse. Any help would be appreciated. 
 A: Embedding is the key to answer your question. For any space $A$ let $C(A)$ denote the $\mathbb{R}$-algebra of continuous bounded real-valued functions on $A$. We can endow it with the $\sup$-norm, but that is irrelevant here.
The embedding $g : X \to Y$ induces the "restriction" $g^\ast : C(Y) \to C(X), g^\ast(\varphi) = \varphi \circ g$ which is a homomorphism of algebras. Define
$$C(X,g) = g^\ast(C(Y)). $$
This is the set of continuous bounded real-valued functions on $X$ which "extend" to $Y$. Since $g(X)$ is dense in $Y$, extensions are unique so that $g^\ast : C(Y) \to C(X,g)$ is an isomorphism. Let $P = \mathbb{R}^{C(X,g)}$ denote the product of copies $\mathbb{R}_\psi$ of the real line indexed by the set of all $\psi \in C(X,g)$. Define
$$u_g : Y \to P, u_g(y) = (\psi'(y))_{\psi \in C(X,g)} ,$$
where $g^\ast(\psi') =\psi$. Then $u_g$ is continuous since $p_\psi \circ u_g  = \psi'$ is continuous for all projections $p_\psi : P \to \mathbb{R}_\psi = \mathbb{R}$. Moreover $u_g$ is an injection because for any two distinct $y_1, y_2 \in Y$ there is a continuous $\varphi : Y \to \mathbb{R}$ such that $\varphi(y_1 ) \ne \varphi(y_2)$ (now take $\psi = g^\ast(\varphi)$). Therefore $u_g$ is an embedding because $Y$ is compact. By construction
$$g' = u_g \circ g : X \to P$$
is an embedding such that $\overline{g'(X)} = u_g(Y)$. For $x \in X$ we obtain
$$g'(x) = u(g(x)) = (\psi'(g(x)))_{\psi \in C(X,g)} = (\psi(x))_{\psi \in C(X,g)} .$$
Now consider $h : X \to Z$. By assumption we have $C(X,g) = C(X,h)$. We therefore obtain an embedding $u_h : Z \to P$ such that $\overline{h'(X)} = u_h(Z)$, where $h' = u_h \circ h$. But $g'= h'$ so that $u_g(Y) = u_h(Z)$ and $\pi = u_h^{-1} \circ u_g : Y \to Z$ is the desired homeomorphism.
