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(This is an incomplete answer. I include it because I think the examples and counterexamples therein may be useful.)

First of all note a somewhat tricky point: the condition (2) cannot be weakened to $\text{Hom}(X,X) = 1$. A counterexample can be found in the ccc of (directed multi)graphs $\text{Hom}((n \rightrightarrows e)^{\text{op}}, \text{Set})$.<sup>1</sup> For let's define $G_1 \triangleq a → b$; this graph has only one homomorphism to itself, but nevertheless there are two different homomorphisms from the graph $G_2 \triangleq a$ to $G_1$.<sup>2</sup> <sup>3</sup>

And, as the OP indicates, we can prove the direction $(1) \Rightarrow (2)$ by showing that $\text{Hom}(1,X)$ is either $\mathbf{1}$ or $\mathbf{0}$, so $\text{Hom}(X×1,X)$ has to be either $\mathbf{1}$ or $\mathbf{0}$, but we can rule out the $\mathbf{0}$ possibility by exhibiting an element in $X^X$: $\text{id} \in \text{Hom}(1×X,X)$ and so $\text{curry(id)} \in \text{Hom(1,$X^X$)}$, so $\text{Hom}(X,X)$ = $\mathbf{1}$ = $\text{Hom}(1,X^X)$. Thus $X^X$ is uniquely isomorphic to $1$.

Unfortunately it is not yet clear to me how to prove the question in the OP directly. I initially thought of doing a case-analysis proof, but clearly there is a large enough variety of objects $X$ such that $X → 1$ is a monomorphism that that approach doesn't seem at all practicable.

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<sup>1</sup> We know that this is cc because all categories of presheaves are cc. For more examples of such categories Reyes et al.'s [*Generic Figures*](https://reyes-reyes.com/wp-content/uploads/2004/06/generic-figures.pdf) is a very readable and interesting source.

<sup>2</sup>By the way $G_1$ and $G_2$ are the representable presheaves for the edge ($よ(e) = \text{Hom(-,e)}$) and for the node ($よ(n) = \text{Hom(-,n)}$), respectively, taken over the category $n \rightrightarrows e$ of which the graphs form a presheaf category.

<sup>3</sup> And indeed we find that, even though $\text{Hom}(1,G_1^{G_1}) = 1$, $G_1^{G_1}$ is quite different from $1$: $\text{Hom}(1,G_1^{G_1}) = 1$ means that $G_1^{G_1}$ has exactly one *loop*, but it turns out to have four nodes and four edges, as shown below!

![The graph $G_1^{G_1}$](https://math.codidact.com/uploads/f57akr1tf1der0g7p9t3urd8pexi)

This is because the nodes of $B^A$ (for any graphs $A$ and $B$) correspond to the elements of $\text{Hom}(よ(n),B^A) = \text{Hom}(よ(n) × A,B)$, and likewise the edges of $B^A$ correspond to the elements of $\text{Hom}(よ(e),B^A) = \text{Hom}(よ(e) × A,B)$. The $\text{source}$ and $\text{target}$ mappings from edges to nodes are defined by the Yoneda embedding: $\text{source} = \text{Hom}(よ(s) × A,B)$, $\text{target} = \text{Hom}(よ(t) × A,B)$ where $s$ and $t$ are the morphisms in the source category $n \rightrightarrows e$.