$\DeclareMathOperator{\Hom}{Hom}$
In a category C, an exponential object $X^Y$ (if it exists) is an object of C such that (by def)
$$\forall Z, \Hom_C(Z,X^Y) \cong \Hom_C(Z \times Y,X)$$
This of course implies that
$$\Hom_C(1,X^Y) \cong \Hom_C(Y,X)$$
i.e. the generalized elements of the exponential object are isomorphic to the hom set.
Is the reverse true? That is, is an object H such that
$$\Hom_C(1,H) \cong \Hom_C(Y,X)$$
always (isomorphic to) the exponential object?
If it's false (as i suspect):
What if one wanted to work in a category with this property?
Under which additional conditions it's true, and/or how do you call a category in which it's true?
(Some context about why i care:
in the type theory view of category theory in which morphisms are
computations, one obviously wants some kind of internalization of homs. But, the full definition of exponential objects looks needlessly strong, I wonder if there is a more fundamental one?)
CORRECTION (edited): As an answer below points out, the definition of exponential object requires a Natural isomorphism. So the first isomorphism is natural.
Best Answer
The claim you ask for is extremely restrictive and I wouldn't expect there to be any nice natural conditions on a category that would make it true. In particular, keep in mind that when you write $$\operatorname{Hom}_C(1,H) \cong \operatorname{Hom}_C(Y,X)$$ this is merely an isomorphism of sets. That is, you are saying that if $H$ is a set such that the cardinality of the set of homomorphisms $1\to H$ is the same as the cardinality of the set of homomorphisms $Y\to X$, then $H$ admits the structure of an exponential object $X^Y$. In the case where $Y=1$, this says in particular that an object $X$ of $C$ is determined up to isomorphism by merely the number of homomorphisms $1\to X$. This is virtually never true.
For an explicit example where this fails but exponential objects exist, consider $C=\mathtt{Set}\times\mathtt{Set}$. Then, for instance, the objects $(1,\mathbb{N})$ and $(\mathbb{N},\mathbb{N})$ both have $\aleph_0$ homomorphisms from the terminal object $1=(1,1)$, but they are not isomorphic.
(Note that even asking for an isomorphism of sets $\operatorname{Hom}_C(Z,H) \cong \operatorname{Hom}_C(Z \times Y,X)$ for all $Z$ is not enough to conclude that $H$ is an exponential object $X^Y$. This isomorphisms also have to be natural in $Z$. For instance, in the full subcategory of $\mathtt{Set}$ consisting of just a single infinite set $X$, there is a bijection $\operatorname{Hom}(Z,X) \cong \operatorname{Hom}(Z \times X,X)$ for all $Z$ (since $X$ itself can be given the structure of a product $X\times X$) but this cannot be made natural in $Z$ so $X$ cannot be made into an exponential object $X^X$.)