Let $X$ be a noncompact $KC$-Space, i.e. a topological space in which every compact set is closed.
Call further $\iota_Y\colon X\to Y$ a compactification if $Y$ is compact and $f$ is an embedding with dense image. Lastly, let $X^\ast := X\cup \{∞\}$ denote the one-point compactification of $X$, with corresponding embedding $\iota_\ast$. (Note that since $X$ is not necessarily Hausdorff nor locally compact, $X^\ast$ does not have to be Haussdorff, which is why we weakened the definition of “compactification” accordingly)
Question: Is there a unique continuous map $h\colon Y\to X^\ast$ which respects the embeddings, i.e. for which $h\circ \iota_Y = \iota_\ast$?
Or, in other words: If $\mathscr C$ is the category with objects $(Y,\iota_Y)$ as above, and morphisms continuous inclusion-respecting maps $Y\to Y^\prime$, is $(X^\ast, \iota_\ast)$ a terminal object?
Note that if $h$ is continuous, it must be surjective: Since $h$ commutes with the embeddings, we have $$
X^\ast\setminus\{∞\} = \mathrm{Im}(\iota_\ast)\subseteq \mathrm{Im}(h) \subseteq X^\ast.
$$ however, since $\mathrm{Im}(h)$ is the image of a compact set and hence compact itself, but $\mathrm{Im}(\iota_\ast) \simeq X$ is noncompact, only $\mathrm{Im}(h)=X^\ast$ remains a possibility.
I've managed to show that – assuming $h$ sends all the new points in $Y$ to $∞$ – neighborhoods around $\iota_\ast(x)\in X^\ast$ have a neighborhood as a preimage, and that preimages of open neighborhoods around $∞$ are open if and only if every compact closed $C\subseteq X$ maps to a closed $\iota_Y(C)$ – but I don't see how that's necessarily the case since $h$ does not have to be closed.
However, I didn't manage to show that all such $h$ necessarily have to map all „new“ points (those in $Y\setminus \mathrm{Im}(\iota_Y)$) to $∞$.
Best Answer
[For convenience, I will pretend in this answer that any compactification actually contains $X$ as a subspace, so I don't have to constantly be writing down the embedding maps.]
No. For instance, you can define a compactification $Y=X\cup\{\infty\}$ where the only neighborhood of $\infty$ is the entire space (and every open subset of $X$ remains open). There will not exist any morphism from this compactification to $X^*$ unless the topology on $X^*$ happens to be the same as the topology on $Y$ (i.e., the only compact closed subset of $X$ is the empty set; given your assumption that $X$ is KC and noncompact, this is impossible!).
A separate issue is that $X$ is open in $X^*$, so if you have some other compactification in which $X$ is not open, you cannot expect it to have a morphism to $X^*$. There are also uniqueness issues--a morphism to $X^*$ does not need to send all the new points to $\infty$ (for instance, if $X$ is uncountable with the cocountable topology, you could let $X'$ be $X$ together with one more point with the cocountable topology and let $Y$ be the 1-point compactification of $X'$, and then the new point of $X'$ can map to anywhere in $X^*$ and the map will still be continuous). With non-Hausdorff spaces, a continuous map is not determined by its values on a dense subset, so generally it will be very hard to get any sort of uniqueness property like this without stronger hypotheses.
If you restrict your definition of "compactification" to require $Y$ to also be KC and that $X$ is open in $Y$, then it is true that $X^*$ is the terminal compactification (assuming $X^*$ is a compactification at all by this definition--it won't always be KC). These hypotheses make it trivial to check that the map $Y\to X^*$ sending every new point to $\infty$ is continuous (the hypothesis that $X$ is open in $Y$ gives continuity at points of $X$, and the hypothesis that $Y$ is KC gives continuity at new points).
For uniqueness, suppose $h:Y\to X^*$ is a morphism of compactifications, and let $A=X\cup h^{-1}(\{\infty\})\subseteq Y$. Then I claim $A$ is compact. To prove this, note that $h^{-1}(\{\infty\})$ is closed in $Y$ and hence compact, so it suffices to show any ultrafilter $F$ on $X$ has a limit in $A$. By compactness of $Y$, $F$ has a limit $y\in Y$; if $y\in A$ we're done, so we may assume $y\not\in A$. In that case $h(y)\neq\infty$, so it is a point of $X$, and then since $h$ is the identity on $X$, $F$ must converge to $h(y)$ in $X$. Thus $h(y)$ is a limit of $F$ in $A$.
Thus since $Y$ is KC, $A$ is closed in $Y$. Since $A$ contains $X$ and $X$ is dense in $Y$, this means $A=Y$. Thus $h$ must map every point of $Y\setminus X$ to $\infty$.