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 *part* 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 *arbitrary* 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 $\alpha$.


**Example 1.**  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


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


The partial definition of $\alpha\_\star$ requires $\alpha\_\star(Tf) = t(Tf)$ where $t = \alpha\_\star(\mathrm{id}\_\star)$, for all $f \in \mathcal{S}$. There are only two possibilities for $Tf$. Taking $f \in \{\mathrm{id}\_\star, \sigma, \sigma^2 \}$ gives nothing, whereas taking $f \in \{\tau, \sigma\tau, \sigma^2\tau\}$ gives $\alpha\_\star(\tau) = t\tau$. Thus $t$ determines $\alpha\_\star(\tau)$, but not, for instance, $\alpha\_\star(\sigma)$.


**Example 2.** (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


$$\mathrm{Hom}\_{\mathcal{D}}\bigl( -, \mathrm{Sp}^\mu \bigr),$$


taking values in $\mathcal{C}$, not $\mathrm{Set}$ as above. (There is a $\mathrm{GL}\_d(\mathbb{C})$ action on the hom-set because the $\mathrm{GL}\_d(\mathbb{C})$ action on each $T(U)$ commutes with $S\_r$; the action on $\mathrm{Sp}^\mu$ must be defined somehow to make this module an object in $\mathcal{D}$, so take the trivial action.) 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,


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


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 $\mathrm{Set}$.



Yoneda map for a composition of a representable functor and an arbitrary functor
This property of the functor $T$ is called being a [dense functor](https://ncatlab.org/nlab/show/dense+functor), or "densely generating functor". The notion was first introduced by Isbell in the case where $T$ is fully faithful under the name "adequate subcategory", but that usage has by now disappeared, and "dense" is preferred. It has been occasionally rediscovered -- for example, in the $\infty$-categorical context, Lurie calls such a functor "strongly generating", but even there the term "dense" is now preferred, especially because "strong generator" means something else classically.


One of the equivalent ways to say that a functor $T: C \to D$ is dense is to say that the induced functor $T^\ast: \mathrm{Set}^{D^{op}} \to \mathrm{Set}^{C^{op}}$ defined by $F \mapsto F \circ T^{op}$ on categories of presheaves is fully faithful. The condition as you've stated it says that $T^\ast$ is fully faithful on the homset from the representable $\mathrm{Hom}\_D(-,Y)$ to the functor $F$, but this is equivalent because every presheaf is a colimit of representables such as $\mathrm{Hom}\_D(-,Y)$.


