I gather in an answer what has been said in the comments to the starting post.
The most important issue concerning the boxed question is that it is too general. As we showed in comments an answer strongly depends on the structure we consider and such a general statement allows us to construct completely trivial answers. This answers can be either positive or negative depending whether we put additional conditions the structure has to fulfill or not.
For example, consider the category of topological spaces with morphisms being continuous maps. The question is: given two topological spaces $X$ and $Y$ is there a map $f:X\to Y$? In this general statement the answer is positive with the argument: every constant function is continuous. Ok, what about nonconstant maps? If the space $X$ is connected, $|X|>1$, and $Y$ is totally disconnected (only singletons are connected, eg. $Y$ is the Cantor set), then such a map does not exist, because continuous maps preserve connectedness. So again we can add some additional conditions to the questions, for example we can consider the maps which are embeddings (ie. homeomorphisms onto the image). And again the answer is negative, for we cannot embed the circle $S^1$ into the segment $[0,1]$.
The same concerns algebraic structures. For example, take the category of vector spaces with linear maps as morphisms. For vector spaces $V$ and $W$, there is always a zero map $z:V\to W$ which is of course linear. But this gives us nothing. So let us constrain the linear maps only to injections. Then the answer is negative, because if $\dim V>\dim W$, then there is no injection of $V$ into $W$.
Similarly, if we consider category of fields and maps being homomorphisms, then for two fields of different characteristics there is no homomorphism between them in case $1$ is added to the structure and there is always a zero homomorphism if we remove $1$ from the structure.
In case of measure spaces the answer is similar to the above ones. Let $(X,\mu)$ and $(Y,\nu)$ be two measure spaces. If $\mu(X)>\nu(Y)$, then obviously there is no measurable measure preserving function between $X$ and $Y$. But if $X\subseteq Y$ and $\mu=\nu$, then the identity (the inclusion) simply preserves $\mu$.
And there are many, many other examples like mentioned by Arturo Magidin the category of posets or even the simple category of sets and any functions as morphisms showing that the answer strongly depends on the structure we consider.