Suppose $\mathcal{C}$ is a Cartesian Closed Category. Letting $B^A$ denote the internal hom of $A$ and $B$ and $ev_B^A:B^A\times{}A\to{}B$ denote the evaluation morphism, there is a bifunctor

\begin{align*} E:\mathcal{C}^{op}\times\mathcal{C}&\to\mathcal{C}\\\\ (A,B)&\mapsto{}B^A\\\\ (f:A\stackrel{\mathcal{C}^{op}}{\to}{}A',g:B\to{}B')&\mapsto{}g^f:B^A\to{}(B')^{(A')} \end{align*}

where $g^f$ is the unique morphism such that

$$ev_{B'}^{A'}\circ{}(g^f\times{}1_{A'})=g\circ{}ev_B^A\circ{}(1_{B^A}\times{}f).$$

This functor is analogous to the hom bifunctor in many respects. In particular, we may transpose it to get the analogues to the covariant and contravariant Yoneda embeddings

\begin{align*} \mathcal{C}&\to{}\mathcal{C}^{\mathcal{C}^{op}}\\\\ B&\mapsto{}B^{(-)}\\\\ (g:B\to{}B')&\mapsto{}(g^A:B^A\to{}(B')^A)_{A\in{}Ob(\mathcal{C})} \end{align*}

and

\begin{align*} \mathcal{C}^{op}&\to{}\mathcal{C}^{\mathcal{C}}\\\\ A&\mapsto{}(-)^A\\\\ (f:A\stackrel{\mathcal{C}^{op}}{\to}{}A')&\mapsto{}(B^f:B^A\to{}B^{A'})_{B\in{}Ob(\mathcal{C}^{op})} \end{align*}

Now,

1) Neither of these functors are injective on objects. Consider the complete preorder on a nonempty set $X$ as a category. Now take a single selected element $x_0\in{}X$ as the product and exponential of every pair of objects.

2) The usual methods can be used to show that that each of these functors is faithful.

My question is the following: under what conditions can we conclude these functors are full?

One can attempt to use the usual argument, but I have only been able to show that the (necessarily unique) preimage of a natural transformation recovers the original transformation on $1$-elements (here, $1$ denotes a terminal object). This is because $B^f$ does not act exactly as precomposition, except on $1$-elements. As a result, these functors will be full if $1$ is a separator in $\mathcal{C}$, but this is a strong condition and I would be interested in other criteria.