Consider the linear map $f:\mathbb{R}^{n\times m}\to \mathbb{R}^{n\times 1}$ defined as follows $$ f(X) = X 1_{m\times 1} $$ This is essentially a reduce operation that collapses the rows of $X$ into their sum. What is the adjoint of this operation?
Attempted Solution
$$ \begin{align} \langle X1, y\rangle_{\mathbb{R}^n} &= \sum_{i=1}^n y_i\sum_{j=1}^m x_{ij} \\ &= \sum_i \sum_j y_i x_{ij} \\ &= \text{trace}\left(X^\top y 1_{n\times 1}^\top\right) \\ &= \langle X, y1_{n\times 1}^\top \rangle_F \\ &= \text{vec}(X)^\top \text{vec}(y1_{n\times 1}) \\ &= \langle \text{vec}(X), \text{vec}(y1_{n\times 1}^\top) \rangle_{\mathbb{R}^2} \end{align} $$ still no clue..
You correctly obtained the adjoint in the 4th line of your argument:
$$\langle X\,1_{m\times 1},\,y\rangle_{\Bbb R^n}\ =\ \langle X,\,y\,1_{1\times m}\rangle_{\Bbb R^{n\times m}}$$ so the adjoint of the linear map $f(X)=X\,1_{m\times 1}$ is the linear map $f^*(y)=y\,1_{1\times m}$.
Note that the definitions of both these maps involve multiplying by a fixed matrix from the right, however, if we vectorize $\Bbb R^{n\times m}$, just as for any linear map, we can write them in the form $$f(x)=Ax\quad f^*(y)=By$$ for some matrices $A\in\Bbb R^{n\times nm}$ and $B\in\Bbb R^{nm\times n}$, and we'll also have $B=A^\top$.