No, you can not write this transformation as $c_{i\uparrow}^\dagger|0\rangle=c_{i\downarrow}^\dagger|0\rangle$, because $c_{i\uparrow}^\dagger|0\rangle$ and $c_{i\downarrow}^\dagger|0\rangle$ are two orthogonal quantum states: they can not be equal. The transformation $c_{i\uparrow}^\dagger|0\rangle\to c_{i\uparrow}|0\rangle$ you start with is also wrong, because the resulting state is $c_{i\uparrow}|0\rangle= 0$, which makes the transformation not unitary. What you normally use $c_{i\uparrow}^\dagger=c_{i\downarrow}^\dagger$ is still wrong, because $c_{i\uparrow}^\dagger$ and $c_{i\downarrow}^\dagger$ are different operators and can not be equal.
The correct way of writing everything is to start by defining a particle-hole transformation operator $\mathcal{P}$, such that its action on the fermion operator is
$$\mathcal{P}c_{i\sigma}^\dagger\mathcal{P}^{-1} = c_{i\sigma}, \tag{1}$$
and its action on the vacuum state is
$$\mathcal{P}|0\rangle = \prod_{i,\sigma} c_{i\sigma}^\dagger|0\rangle. \tag{2}$$
It is important that the vacuum state $|0\rangle$ must also transform under $\mathcal{P}$. Physically, it is because the vacuum state is the state with no particle occupation, which means it is a state that is fully occupied by holes. So under the particle-hole transformation, the vacuum state will become a fully occupied state of particles, as expressed in Eq.(2). Mathematically, Eq.(2) follows from Eq.(1) as a result of consistency. Because the vacuum state is defined as the state annihilated by all the annihilation operator, i.e. $c_{i\sigma}|0\rangle=0$. Now if we apply $\mathcal{P}$ to both sides of the equation, we have $\mathcal{P}c_{i\sigma}|0\rangle=\mathcal{P}0=0$ (because any linear operator acts on $0$ is still $0$). However the left-hand-side reads $\mathcal{P}c_{i\sigma}|0\rangle=\mathcal{P}c_{i\sigma}\mathcal{P}^{-1}\mathcal{P}|0\rangle=c_{i\sigma}^\dagger\mathcal{P}|0\rangle$, meaning that the state $\mathcal{P}|0\rangle$ is annihilated by creation operator $c_{i\sigma}^\dagger$ instead, so the state $\mathcal{P}|0\rangle$ has to be a fully occupied state.
Applying Eq.(1) and Eq.(2) to a single site (omitting the site index $i$), we have
$$\mathcal{P}c_{\uparrow}^\dagger|0\rangle = (\mathcal{P}c_{\uparrow}^\dagger\mathcal{P}^{-1})(\mathcal{P}|0\rangle)=c_{\uparrow}c_{\uparrow}^\dagger c_{\downarrow}^\dagger|0\rangle=c_{\downarrow}^\dagger|0\rangle.$$
So this means under the particle-hole transformation, the spin-up state $c_{\uparrow}^\dagger|0\rangle$ is transformed to a spin-down state $c_{\downarrow}^\dagger|0\rangle$.
