No, since $\hom(X,Y)$ is a set, but $Y^X$ is an object of $C$.
The relation between these two guys is given by the following formula (where $1$ denotes a terminal object):
$$\hom(1,Y^X) \cong \hom(X,Y).$$
This bijection follows immediately from the definitions.
In other words, $Y^X$ is an object whose set of "global sections" is $\hom(X,Y)$. This is true verbatim if $C$ is a Grothendieck topos.
No, since $\hom(X,Y)$ is a set, but $Y^X$ is an object of $C$.
The relation between these two guys is given by the following formula (where $1$ denotes a terminal object): $$\hom(1,Y^X) \cong \hom(X,Y).$$ This bijection follows immediately from the definitions.
In other words, $Y^X$ is an object whose set of "global sections" is $\hom(X,Y)$. This is true verbatim if $C$ is a Grothendieck topos.