Referring to this definition of compactification: A pair $(f,Y)$ where $f:X\longrightarrow Y$ is an embedding, $Y$ is compact and $\overline{f(X)}=Y$, is called a compactification of a topological space $X$.
If $X$ is compact, it is clear that $X$ is its compactification.
Now, for the Stone-$\check{C}$ech compactification $\beta X$ of $X$: If $X$ is compact and Hausdorff, then $X$ is the same as $\beta X$ (I am using the fact that to topologists, homeomorphic spaces are the same).
For the Wallman compactification $wX$ of $X$: If $X$ is compact and Hausdorff, then $X$ is normal, making $wX=\beta X$. Thus $X$ is the same as $wX$.
I don't see a compact space $X$ being homeomorphic to its compactification unless that compactification is Hausdorff: Say $(f,Y)$ is a compactification of a compact Hausdorff space $X$, we have that $f(X)$ is compact in $Y$. There is no guarantee that $f(X)=Y$. Unlike when $Y$ is also Hausdorff, then $f(X)$ will definitely be equal to $Y$.
The preceding paragraph is my main concern. Is my assumption true?
(Excuse me for abusing the word compactification, especially for an already compact space.)
Take $X$ a single point and $Y$ two points, both spaces given the indescrete topology. Let $f: X \to Y$ be an arbitrary map (which will be an embedding). $X$ is compact and $\overline{f(X)} = Y$ but it's not the case that $X$ is homeomorphic to $Y$.