$\newcommand{\Ord}{\text{Ord}}
\newcommand{\ZFC}{\text{ZFC}}$
Here is one way to formalize your concept a little more tightly,
which provides the answers to your questions. For any large
cardinal property $P$, let's take the phrase "$\Ord$ is $P$" to
be the theory asserting $\sigma$, for any sentence $\sigma$ that
ZFC proves is true in $V_\delta$ under the assumption that
$\delta$ has property $P$ in $V$.
With this formalization, "$\Ord$ is $P$" asserts that the universe is just like we would expect, if we were living inside $V_\delta$ for an actual $P$ cardinal $\delta$. For example, "$\Ord$ is measurable" implies that there are a proper
class of weakly compact cardinals, since this is true in
$V_\delta$, whenever $\delta$ is measurable, and "$\Ord$ is
supercompact" implies that there are many partially supercompact
cardinals, with nice limit properties. For example, they would
form a stationary class in the sense that every definable class
club would contain one of them. This notion seems to capture what
one would want to mean by saying $\Ord$ has property $P$ as a
purely first-order theory about sets.
With this idea, the point I would like to make is that assuming
$\Ord$ is $P$ is essentially equivalent to assuming that what you
have is $V_\delta$, where $\delta$ has property $P$ in a larger
universe.
Theorem. For any large cardinal property $P$, a model of set
theory $M$ satisfies "$\Ord$ is $P$" if and only if $M\prec
V_\delta^N$ for some taller model of set theory $N$ with a
cardinal $\delta$ having property $P$ in $N$.
Proof. The backward direction is immediate, since $\delta$ having
property $P$ implies that $V_\delta$ satisfies every assertion of
$\Ord$ is $P$. For the forward direction, suppose $M$ satisfies
$\Ord$ is $P$. Let $T$ be the theory consisting of $\ZFC$, plus the
assertion "$\delta$ is $P$", using a new constant symbol $\delta$,
plus the assertions $\varphi^{V_\delta}$, for any $\varphi$ in the
elementary diagram of $M$, using constants for elements of $M$.
This theory is finitely consistent, since otherwise there would be
finitely many assertions in it that are contradictory, and so
there would be a statement $\varphi$ true in $M$ that provably
could not hold in $V_\delta$ for any cardinal $\delta$ with
property $P$. But that would contradict our assumption that
$M\models\Ord$ is $P$.
If $N$ is any model of the theory, then $\delta$ has property $P$
in $N$, and we get $M\prec V_\delta^N$, because $V_\delta^N$
satisfies the elementary diagram of $M$. Another way to say this
is that there is an elementary embedding $j:M\to V_\delta^N$,
mapping every element of $M$ to the interpretation of its constant
in $N$. QED
Thus, if one is inclined to assume $\Ord$ is $P$, then why not go ahead and make the full move to a model with an actual $P$ cardinal $\delta$, such that our old
world looks exactly like $V_\delta$ in this new world. In particular, under this terminology, the theory $\ZFC+\Ord$ is $P$ is equiconsistent with $\ZFC+\exists \delta$ with property $P$.
Corollary. The following theories are equiconsistent:
- $\ZFC+\Ord$ is $P$.
- $\ZFC+\exists \kappa$ with property $P$.
Lastly, I would like to point out that there is some variance in
the literature about what "$\Ord$ is $P$" should mean. For
example, one often finds the phrase "$\Ord$ is Mahlo" to mean only
the weaker assertion, that every definable closed unbounded class
of cardinals contains a regular cardinal. This is what one finds, for example,
at Cantor's Attic. But this is strictly weaker
in consistency strength than ZFC+$\exists\kappa$ Mahlo, since this latter theory implies
the consistency of the former, as it is true in $V_\kappa$
whenever $\kappa$ is Mahlo.
The answer to your question is (almost) yes (almost is because of the addition of DC to the statement).
Recently Gabriel Goldberg has proved
''Con(NBG+DC+Reinhardt)$ \implies$ Con(ZFC+I0)''.
See the abstract of the talk by Gabriel Goldberg Choiceless cardinals and I0.
(Thanks to Rahman for pointing this to me).
Edit. The result of Goldberg is now available,where indeed something stronger is proved. See Even ordinals and the Kunen inconsistency. It is shown, assuming DC, the existence of an elementary embedding from $V_{λ+3}$ to $V_{λ+3}$ implies the consistency of ZFC + $I_0$, while by a recent result of Schlutzenberg, an elementary embedding from $V_{λ+2}$ to $V_{λ+2}$ does not suffice.
The paper of Schlutzenberg is
On the consistency of ZF with an elementary embedding $j : V_{λ+2} → V_{λ+2}$.
Best Answer
The usual relations to consider in the large cardinal hierarchy are
Your concept, however, is focused on the least instance of the large cardinal notion, and this is also studied.
In broad terms, the large cardinal hierarchy is roughly linear, with the stronger cardinals being stronger with respect to all three of these relations. In most instances, we have that every A cardinal (the stronger notion) is also a B cardinal, as well as a limit of B cardinals, and so we get also the consistency implication and the least A cardinal is strictly larger than the least B cardinal.
However, there are some notable deviations from this. These deviations come in two types.
First, there are the instances where a large cardinal concept A has stronger consistency strength than B, but the least instance of A is definitely less than the least instance of B. For example, a superstrong cardinal has higher consistency strength than a mere strong cardinal, since if $\kappa$ is superstrong, then $V_\kappa\models$ ZFC + there is a proper class of strong cardinals, but the least superstrong cardinal is definitely less than the least strong cardinal. This is simply because superstrongness is witnessed by a single object, and strong cardinals are $\Sigma_2$ reflecting, and therefore reflect the least instance below.
There are numerous similar instances of this. Any time a large cardinal notion is witnessed by a single object or is witnessed inside some $V_\theta$ — and this would include weakly compact, Ramsey, measurable, superstrong, almost huge, huge, rank-to-rank and others — then the least instance of that cardinal will be less than the least $\Sigma_2$-reflecting cardinal and indeed less than the least $\Sigma_2$-correct cardinal. But $\Sigma_2$ correct cardinals provably exist in ZFC, and therefore have very low consistency strength.
So we have numerous interesting instances where your $<$ order does not align with consistency strength:
Meanwhile, second, there are examples of your $\perp$ situation, where the size of the smallest instance is not yet settled. This phenomenon is known as the "identity-crises" phenomenon, named by Magidor when he proved that the least measurable can be the same as the least strongly compact, or strictly less, depending on the model of set theory. Many further instances of this are now known, some of which appear in my paper:
This paper provides many instances of your $\perp$ situation, where the question of whether the least A cardinal is smaller than or the same size as the least B cardinal is not settled in ZFC.
Finally, let me qualify my remark that the large cardinal hierarchy is roughly linear. The hierarchy is indeed mainly linear, but one sometimes hears stronger assertions of linearity, as something that we know and which needs explanation, but I don't feel these knowledge claims are justified. Of course, the identity crises phenomenon provides instances of non-linearity in the direct implication hierarchy, and so when large cardinal set theorists assert that the large cardinal hierarchy is linear, they are speaking of the consistency strength order. So let me mention a few cases where we simply don't yet know linearity:
A supercompact cardinal versus a strongly compact plus an inaccessible above.
A supercompact cardinal versus a proper class of strongly compact cardinals.
A Laver-indestructible weakly compact cardinals versus a strongly compact cardinal.
A cardinal $\kappa$ that is $\kappa^+$-supercompact versus $\kappa$ is $\kappa^{++}$-strongly compact.
A PFA cardinal versus a strongly compact cardinal.
And many others.
My perspective is this. Because we have essentially no method for proving non-linearity in the consistency strength hierarchy, it is not surprising that we see only instances of linearity, and this may be a case of confirmation bias. But don't get me wrong: of course I agree that the consistency strength hierarchy is mainly linear in broad strokes.