= Solution
<Nash's theorem> says that every finite game, including a two-player game with two pure choices each, has at least one <Nash equilibrium> in <mixed strategies>.
For player $1$, holding $p_2$ fixed, the <expected value> of the payoff is
$$
E_1=-Up_1p_2+2p_1(1-p_2)-2(1-p_1)p_2+(1-p_1)(1-p_2)
=1-3p_2+p_1\bigl[1-(U-1)p_2\bigr].
$$
It is an <affine function> of $p_1$. Therefore player $1$ chooses $p_1=1$ when $1-(U-1)p_2>0$, chooses $p_1=0$ when it is negative, and is indifferent when it vanishes. By <symmetry>, player $2$ has the analogous <best response>.
If $U<2$, then $1-(U-1)p>0$ for every $p\in[0,1]$, so becoming a Dark Lord with probability one is a <strictly dominant strategy>. The unique equilibrium is
$$
\boxed{(p_1,p_2)=(1,1).}
$$
If $U>2$, each player is indifferent when the other chooses $p=1/(U-1)$. This gives one interior mixed equilibrium, while the two asymmetric pure equilibria arise because a player facing a certain peasant chooses Lord and a player facing a certain Lord chooses peasant. The three equilibria are
$$
\boxed{(1,0),\qquad(0,1),\qquad
\left(\frac1{U-1},\frac1{U-1}\right).}
$$
Consequently $\boxed{U_0=2}$. At the threshold itself there is a continuum of equilibria with $p_1=1$ or $p_2=1$, consistent with the question's strict inequalities.
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