Consider $\mathrm{AD}_X$, determinacy for games where players pick moves from $X$. We know that it is consistent for $X = \omega$ or $\mathbb{R}$ (under large cardinal assumptions), but inconsistent for $X = \omega_1$.

Since this implies determinacy is inconsistent for any set with $\omega_1 \leq X$, this answers the question of consistency for most cardinalities that we might consider (e.g. $\omega_1 \cup \mathbb{R}$, $\mathcal{P}(\mathbb{R})$, all uncountable ordinals). Of course, in the absence of choice other cardinalities might exist. This leads me to the following two questions:

Q1:Is the existence of infinite Dedekind finite sets consistent with $\mathrm{AD}$?

Q2:If so, is 'there exists an infinite Dedekind finite set $X$ such that $\mathrm{AD}_X$ holds' consistent? (Let's call this statement $\mathrm{AD}_\mathrm{DF}$.)

(Note that assuming the consistency of $\mathrm{AD_\mathbb{R}}+\mathrm{DC}$, we have that $\mathrm{AD}_\mathrm{DF}$ can't be a consequence of either $\mathrm{AD}$ or $\mathrm{AD_\mathbb{R}}$.)

Games on infinite Dedekind finite sets seem very strange. For example, consider the game where the first player to play a move that's already been played loses. By Dedekind finiteness somebody has to win this game (and in finite time!), but both players always have not-losing moves at every point. I initially thought this game would be nondetermined, but someone pointed out the following to me:

Let $X$ be infinite Dedekind finite. Then $2 \times X$ is as well. Player 2 has a winning strategy for this game on $2 \times X$: given that an odd number of moves have been played, if nobody has lost yet then there must be some $x \in X$ such that one but not both of $(0, x)$ and $(1, x)$ have been played. Thus we play the other one.

(This means that Player 1 has a winning strategy for this game on $2 \times X \setminus (0, x)$, by pretending they are Player 2 in the game on $2 \times X$ and that the first move was $(0, x)$.)