Let $\mathcal{C}$ and $\mathcal{D}$ be categories and let $T : \mathcal{C} \rightarrow \mathcal{D}$ be a functor. Suppose that $F : \mathcal{D}^\mathrm{op} \rightarrow \mathrm{Set}$ is a functor. (So in the older language I am more used to, $F$ is a contravariant functor.) Let $Y \in \mathcal{D}$ be an object. Under what conditions on $T$ and $Y$ is there a bijection
$$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl(T(-), Y\bigr), F\circ T\Bigr) \stackrel{\mathrm{Yon}}{\longrightarrow} F\bigl(Y\bigr)$$
with the properties of the Yoneda embedding?

The rest of my question is background and motivation. I'll first show that there is a unique way to define <em>part</em> of $\mathrm{Yon}$ when $Y$ is $T(B)$ for some object $B \in \mathcal{C}$. Suppose that $\alpha$ is a natural transformation from the contravariant functor $\mathrm{Hom}_\mathcal{D}\bigl(T(-), T(B) \bigr) : \mathcal{C} \rightarrow \mathrm{Set}$ to the contravariant functor $F \circ T : \mathcal{C} \rightarrow \mathrm{Set}$. (Again these become ordinary functors replacing $\mathcal{C}$ with $\mathcal{C}^\mathrm{op}$.) Apply 

$$\alpha_{B} : \mathrm{Hom}_{\mathcal{D}}\bigl(T(B),T(B) \bigr) \rightarrow F\bigl(T(B)\bigr)$$ 

to $\mathrm{id}_{T(B)}$ to get $t = \alpha_B(\mathrm{id}_{T(B)}) \in F(TB) = F(Y)$. Then, as in the usual proof of the contravariant version of Yoneda's Lemma, a chase around a commutative square shows that if $f : A \rightarrow B$ is any morphism in $\mathcal{C}$, and so $Tf : T(A) \rightarrow T(B)$ is a morphism in $\mathcal{D}$, then $\alpha_A : \mathrm{Hom}_\mathcal{D}\bigl(T(A),T(B)\bigr) \rightarrow F\bigl(T(A)\bigr)$ satisfies

$$\alpha_A(Tf) = (FTf) \alpha_B(\mathrm{id}_{T(B)}) = (FTf)(t) \in F\bigl(T(A)\bigr).$$

This determines $\alpha_A$ on those morphisms of the form $Tf$, for an <em>arbitrary</em> object $A$ of $\mathcal{C}$, and so when $T$ is full, we have the required bijection. Restated in terms of $t$, my question asks for a necessary and sufficient condition on $T : \mathcal{C} \rightarrow \mathcal{D}$ and $Y \in \mathcal{D}$ for $t \in F(Y)$ to determine $\alpha$. 


<b>Example 1. </b> Let $\mathcal{C} = \mathcal{D}$ be the category with one object $\star$ and $\mathrm{Mor}(\star,\star) = \mathcal{S}$ where $\mathcal{S} = \langle \sigma, \tau \rangle$ is the group of permutations of $\{1,2,3\}$, generated by $\sigma = (1,2,3)$ and $\tau = (1,2)$. Since $Y = \star = T(\star)$, the requirement on $Y$ for part of $\mathrm{Yon}$ to be defined is obviously satisfied. Define $T : \mathcal{C} \rightarrow \mathcal{C}$ by abelianization with embedding $\langle 1, \tau \rangle$, so 

$$T(\mathrm{id}_\star) = T(\sigma) = T(\sigma^2) = \mathrm{id}_\star, \quad T(\tau) = T(\sigma\tau) = T(\sigma^2 \tau) = \tau.$$ 

There are only two possibilities for $Tf$ in my partial definition of $\alpha_\star$. Taking $f \in  \{\mathrm{id}_\star, \sigma, \sigma^2 \}$ gives nothing, whereas taking $f \in \{\tau, \sigma\tau, \sigma^2\tau\}$ gives $\alpha_\star(h)\tau$ for each $h \in \mathcal{S}$. Thus $\alpha$ is a natural transformation if and only  $\alpha_\star(\sigma^i \tau) = \alpha_\star(\sigma^i) \tau$ for each $i$. So we must have $\alpha_\star(\tau) = t\tau$, but $t$ does not determine $\alpha_\star(\sigma\tau) = \alpha_\star(\sigma)\tau$ or $\alpha_\star(\sigma^2\tau) = \alpha_\star(\sigma^2)\tau$.

<b>Example 2.</b> (Edited, since although this was the original motivation, I realised later it doesn't exactly fit the setup above.) Let $\mathcal{C}$ be the category of representations of the algebraic group $\mathrm{GL}_d(\mathbb{C})$ and let $\mathcal{D}$ be the category of bimodules with left $\mathrm{GL}_d(\mathbb{C})$ action and right $S_r$ action. Let $T : \mathcal{C} \rightarrow \mathcal{D}$ be the functor defined by $T(U) = U^{\otimes r}$, where the tensor product is regarded as a representation of $S_r$ acting on the right by place permutation on tensors. Let $F$ be the representable functor 

$$\mathrm{Hom}_{\mathbb{C}S_r}\bigl( -, \mathrm{Sp}^\mu \bigr),$$ 

taking values in $\mathcal{C}$, not $\mathrm{Set}$ as above. (There is a $\mathrm{GL}_d(\mathbb{C})$ action because its action on each $T(U)$ commutes with $S_r$.) By Schur–Weyl duality, 

$$U^{\otimes r} \cong \bigoplus_\nu \mathrm{\Delta}^\nu(U) \boxtimes \mathrm{Sp}^\lambda,$$

where $\Delta^\nu$ is the Schur functor for $\nu$. Hence by Schur's Lemma, $F\bigl(T(U)\bigr) \cong \Delta^\mu(U)$, naturally in $U$. Taking $Y = \mathrm{Sp}^\lambda$, we have, 

$$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\lambda\bigr), \mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\mu\bigr)) \cong \mathrm{Nat}(\Delta^\lambda, \Delta^\mu) $$

and by further applications of Schur's Lemma,

$$ \mathrm{Nat}(\Delta^\lambda, \Delta^\mu) \cong \begin{cases} \mathbb{C} & \text{if $\lambda=\mu$} \\ 0 & \text{otherwise} \end{cases} \cong \mathrm{Hom}_{\mathrm{C}S_r}(\mathrm{Sp}^\lambda, \mathrm{Sp}^\mu) \cong F(\mathrm{Sp}^\lambda)$$

so every natural transformation comes from an element of $F(\mathrm{Sp}^\lambda) = F(Y)$, as in the Yoneda embedding, but now into the category $\mathcal{C}$ of representations of $\mathrm{GL}_d(\mathbb{C})$, rather than $\mathrm{Set}$.