In this answer, solving an exercise of Halmos, it is shown that:
If $x$ and $y$ are non-zero vectors and $\langle x,y\rangle$ is (real and) strictly positive, then there exists a positive semi-definite transformation $A$ such that $Ax=y$.
The answer is by construction. Letting $P$ be the orthogonal projection on the subspace spanned by $y$, we can choose $$ A=\frac{\Vert y\Vert^2}{\langle x,y\rangle} P. $$
This works, but leaves me slightly disappointed, since the map is only semi-definite. My gut-feeling says that, surely, a positive (non-semi-)definite transformation should exist as well... is that true? And can we point to a simple one like the one given above?
As you suspected, the answer is a resounding yes!
I will use an inner product space over $\Bbb R$; the case over $\Bbb C$ is similar.
In the simple case when $x=\pmatrix{1\\0}$ and $y=\pmatrix{a\\b}$ where $a>0$, we can let $A=\pmatrix{a& b\\b&\frac{1+b^2}a}$, which is a positive-definite symmetric matrix such that $Ax=y$. The general case can be reduced to the simple case by an orthogonal transformation followed by a scalar multiplication.
A more geometrical approach is the following.
Let $\mathcal V$ be the ambient vector space (which is isomorphic to $\Bbb R^n$ for some $n$). There are two cases.
$x$ and $y$ are linearly dependent.
Suppose $y=ax$ for some $a\in\Bbb R$. Then $$a= \frac{\langle ax, y\rangle}{\langle x, y\rangle}=\frac{\|y\|^2}{\langle x, y\rangle} > 0.$$ Let $Au=au$ for all $u\in U$.
Then $A$ is a positive-definite transformation such that $Ax=y$.
$x$ and $y$ are linearly independent.
Let $\mathcal U$ be the subspace spanned by $x$ and $y$.
Since the angle between $x$ and $y$ is less than the right angle, we can find vector $s$ and vector $t$ in $\mathcal U$ such that the angle between $s$ and $t$ is a right angle with $x,y$ strictly inside. That means, $\langle s,t\rangle=0$ and $\langle x,s\rangle>0$, $\langle x,t\rangle>0$, $\langle y,s\rangle>0$, $\langle y,t\rangle>0$.
Since $s$ and $t$ span $\mathcal U$, we can express $x$ and $y$ as linear combinations of $s$ and $t$. Suppose $$\begin{aligned} x = x_s s + x_t t\\ y = y_s s + y_t t\end{aligned}$$ where $x_s, y_s, x_t, y_t$ are positive real numbers.
Let $A$ be the linear transformation such that $$\begin{aligned} As&=\frac{y_s}{x_s}s,\\At&=\frac{y_t}{x_t}t,\\Av&=v\quad\text{for }v\in \mathcal U^\perp \end{aligned}$$ It is straightforward to verify that $A$ is positive-definite transformation such that $Ax=y$.