In which category an object isomorphic to the hom set is the exponential object? $\DeclareMathOperator{\Hom}{Hom}$
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.
 A: 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 $X^X$.)
A: This is not true.
Consider $A$, which is the poset $\{0, 1\}$ with partial order $\leq$ expressed as a category.
Consider $C = Set^{A^{op}}$, the category of presheaves on $A$ (that is, the category of functors $A^{op} \to Set$).
Note: if you desire that $C$ be a small category, simply replace $C$ with $FinSet^{A^{op}}$, where $FinSet$ is the full subcategory of $Set$ consisting of sets of the form $\{x \in \mathbb{N} \mid x < m\}$ for some $m$.
Recall that there is a fully faithful functor $Y : A \to Set^{A^{op}}$ given on objects $Y(J)(K) = Hom_A(K, J)$. Note that $Y(1)$ is the terminal object in this case.
Let $X = Y(0)$, and let $W = Y(1) = 1$. Let $Z = 0$ be the functor sending every object to $\emptyset$.
Then $Hom_C(1, Z) = \emptyset = Hom_C(W, X)$. This is because $Hom_C(W, X) = Hom_C(Y(1), Y(0)) \cong Hom_A(1, 0) = \emptyset$.
However, $Z$ is not the exponential object $X^W$. This object is, in fact, $X$. This follows easily from the fact that $W \cong 1$, and therefore $X^W \cong X$.
There are, however, categories in which the property does in fact hold. In particular, the property is equivalent to the claim that if there is a bijection $Hom_C(1, X) \to Hom_C(1, Y)$, then there is an isomorphism $X \to Y$.
The main property which I can think of that guarantees this is true is that the functor $X \mapsto Hom_C(1, X)$ is fully faithful. This is true when $C$ is the category of sets, since here, the functor $X \mapsto Hom_C(1, X)$ is naturally isomorphic to the identity functor.
Let $C$ be a category with finite limits. $1$ is said to be a strong generator when for any monic $m : A \to B$, if for all $x : 1 \to B$, $x$ factors through $m$, then $m$ is an isomorphism. Note that if $1$ is a strong generator and $f, g : A \to B$, then if for all $x : 1 \to A$, $f \circ x = g \circ x$, then $g = f$. This follows by taking the equaliser of $f$ and $g$. So $1$ is a generator when $1$ is a strong generator.
If $1$ is a strong generator, then let us suppose that we have $f : A \to B$ and for all $w, x : 1 \to A$, if $f \circ w = f \circ x$ then $w = x$. Then consider the kernel pair $p_1, p_2 : K \to A$. I claim that $p_1 = p_2$. For consider some $x : 1 \to K$. Then we have $f \circ (p_1 \circ x) = (f \circ p_1) \circ x = (f \circ p_2) \circ x = f \circ (p_1 \circ x)$. Then $p_1 \circ x = p_2 \circ x$. Thus, since $1$ is a generator, $p_1 = p_2$. Then $f$ is monic.
This last result is equivalent to saying that the functor $C \to Set$ sending $X$ to $Hom_C(1, X)$ is faithful.
Now let us assume that $f : X \to Y$ becomes an isomorphism on $Hom_C(1, X) \to Hom_C(1, Y)$. Then $f$ must be monic, since the functor $X \mapsto Hom_C(1, X)$ is faithful. And since $1$ is a strong generator, we see that $f$ must be an isomorphism.
So $1$ being a strong generator is equivalent to $X \mapsto Hom_C(1, X)$ being faithful and reflecting isomorphisms.
We now the condition which $X \mapsto Hom_C(1, X)$ is full. This condition can be expressed as: for any external predicate $\phi$, if for all $x : 1 \to X$, there exists a unique $y : 1 \to Y$ such that $\phi(x, y)$, then there is some $f : Hom_C(X, Y)$ such that for all $x$, $\phi(x, f \circ x)$. So this is the principle that we can define a function $X \to Y$ by giving a "rule" which relates each $x : 1 \to X$ to a unique $y : 1 \to Y$.
This seems like a pretty innocuous condition, but it's actually an incredibly strenuous one because it deeply connects the meta-theory (aka ZFC in most cases) to the theory of the category in a very deep way.
To illustrate this, let's suppose we have some model $V$ of some variant of set theory (some first-order countable extension of Z, say). By the downward Lowenheim-Skolem theorem, there is a countable model $V'$ of the same theory. Take the category of sets of $V'$. Then $V'$ satisfies all the first-order internal statements about category theory that $V$ does. But nonetheless, consider the internal $\mathbb{N} \in V'$, and consider the internal $P(\mathbb{N}) \in V'$. Then there is no internal isomorphism $\mathbb{N} \to P(\mathbb{N})$ by Cantor's diagonalisation argument. But there is an external bijection $Hom(1, \mathbb{N}) \to Hom(1, P(\mathbb{N}))$ because $V'$ is countable, and because both $\mathbb{N}$ and $P(\mathbb{N})$ must have externally infinitely many internal elements, hence must have externally countably infinitely many internal elements.
