Answer to Question 1
Yes. In ZF a “uniformly supercompact’’ cardinal is automatically Reinhardt.
Suppose κ is uniformly supercompact and let

  j : V → M   (crit j = κ)

be the fixed elementary embedding with M transitive and

  ^λM ⊆ M   for every ordinal λ,

i.e. M is closed under all ordinal–length sequences.
Because every subset of an ordinal is such a sequence (identify a set A⊆λ with its characteristic function λ→2), we have

  𝒫(λ) ⊆ M for every ordinal λ, and in particular 𝒫(Ord) ⊆ M. (1)

Coding sets by sets of ordinals. For every set x we build, without using the Axiom of Choice, a set code(x) ⊆ Ord such that x is definable from code(x) and code(x) is uniquely determined by x. One convenient choice is this.

For each transitive well-founded set a consider the pair 〈α,R〉 where α is the ordinal rank ρ(a) and R⊆α×α is the binary relation that, through the Mostowski collapse, is isomorphic to 〈TC(a),∈〉. Among all such relations pick the <_L-least one (lexicographic comparison by Gödel codes). Call this relation R_a ; then R_a ⊆ Ord and the map a ↦ R_a is parameter-free definable in the language of ZF.

Apply this with a = TC({x}). Let code(x)=R_{TC({x})}. Because the assignment is parameter-free definable, (1) implies

  code(x) ∈ M. (2)

Recovery of x from its code is absolute: inside any transitive class containing code(x), take the Mostowski collapse of the unique well-founded extensional relation coded by code(x) and pick the unique member whose transitive closure has that relation as its collapse. Hence x is definable in M from code(x); by (2) we obtain

  x ∈ M.

So every set of V belongs to M, i.e. V ⊆ M, and of course M ⊆ V, whence

  M = V.

Thus j : V → V, and κ is a Reinhardt cardinal. No use of the Axiom of Choice was made anywhere, so the result is provable in ZF.

Answer to Question 2
Because “uniformly supercompact’’ and “Reinhardt’’ coincide in ZF, ZF + “there is a uniformly supercompact cardinal’’ and ZF + “there is a Reinhardt cardinal’’ are equiconsistent. The exact consistency strength of either theory is presently unknown: it is the same open problem as the consistency of Reinhardt cardinals itself.

(Incidentally, once M = V one also sees that the restriction j↾M is an elementary embedding of V into V, answering the subsidiary remark in the question.)