So the large cardinal axioms are for the most part considered to be linearly ordered by consistency strength. For the large cardinals between extendibility and rank-into-rank (i.e. the $n$-huge cardinals), by using a little help from algebra, I have formulated large cardinal axioms where I have no idea about the consistency strength of these axioms.

Suppose that $\gamma$ is a limit ordinal. If $f$ is a function, then define $f\upharpoonright_{\gamma+1}:V_{\gamma+1}\rightarrow V_{\gamma+1}$ be the function defined by $f\upharpoonright_{\gamma+1}(x)=f(x)\cap V_{\gamma}$ for each $x\subseteq V_{\gamma}$. Define $\mathbf{EE}_{\gamma}$ to be the collection of all functions $f:V_{\gamma+1}\rightarrow V_{\gamma+1}$ such that for each $\alpha>\gamma+1$ there is a $\beta$ and an elementary embedding $j:V_{\alpha}\rightarrow V_{\beta}$ with $j\upharpoonright_{\gamma+1}=f$.

Then $\mathbf{EE}_{\gamma}$ can be endowed with a self-distributive operation $*$ defined by $f*g=j(k)\upharpoonright_{\gamma+1}$ where there are ordinals $\gamma\leq \delta<\alpha\leq\beta$ with $j:V_{\alpha}\rightarrow V_{\beta},k:V_{\gamma}\rightarrow V_{\delta}$ and where $j\upharpoonright_{\gamma+1}=f,k\upharpoonright_{\gamma+1}=g$.

If $X$ is a finite self-distributive algebra, then let $\mathrm{LC}(X)$ be the axiom that posits that there is some limit ordinal $\gamma$ such that $\mathbf{EE}_{\gamma}$ has a subalgebra isomorphic to $X$.

For example, if $(X,*)$ is the algebra where $X=\{0,\dots,n\}$ and $x*y=1$ whenever $x\leq y$ and $x*y=y$ whenever $y<x$, then $\mathrm{LC}(X)$ is equivalent to the existence of $n$ extendible cardinals.

The axioms of the form $\mathrm{LC}(X)$ are large cardinal axioms, but there does not seem to be any natural linear pre-ordering on the class of all finite self-distributive algebras $\leq$ where if $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are both consistent, then $X\leq Y$ if and only if $\mathrm{Con}(\mathrm{LC}(X))\rightarrow\mathrm{Con}(\mathrm{LC}(Y)).$ If large cardinal axioms are supposed to be linearly ordered by consistency strength, then why does there seem to be no linear ordering on the axioms of the form $\mathrm{LC}(X)$?

Are the consistent axioms of the form $\mathrm{LC}(X)$ linearly ordered by consistency strength?

Is the collection of consistent axioms of the form $\mathrm{LC}(X)$ well-founded when ordered by consistency strength?

Is the problem of determining whether $\mathrm{LC}(X)$ has greater consistency strength than $\mathrm{LC}(Y)$ computable in polynomial time?

Is the problem of determining whether $\mathrm{LC}(X)$ has consistency strength equal to $\mathrm{LC}(Y)$ computable in polynomial time?

Does there exist a consistent large cardinal axiom of the form $\mathrm{LC}(X)$ which does not follow from the existence of a non-trivial elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$?

If $\mathrm{LC}(X)$ is consistent, then is it also consistent that there exists some cardinal $\lambda$ and a limit ordinal $\gamma$ where $X$ is isomorphic to some subalgebra of $\mathcal{E}_{\lambda}/\equiv^{\gamma}$?

If $\mathrm{LC}(X)$ is consistent, then is $\mathrm{Con}(\mathrm{LC}(X))$ a consequence of the existence of an $n$-huge cardinal for some $n$?

If $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are consistent, then is the conjunction of $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ also consistent? If $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are consistent, then does there exist an algebra $Z$ that contains both $X,Y$ as subalgebras and where $\mathrm{LC}(Z)$ is consistent?

If the axioms of the form $\mathrm{LC}(X)$ are well-ordered by consistency strength and if the problem $\mathrm{Con}(\mathrm{LC}(X))\rightarrow\mathrm{Con}(\mathrm{LC}(Y))$ is computable , then what is the order type of this well-ordered set?

The axioms $\mathrm{LC}(X)$ are strengthenings of the notion of extendibility. I would also like to eventually see other large cardinal notions generalized to their $\mathrm{LC}(X)$ versions and their consistency strength levels compared so that we can have a much finer picture of the large cardinals between extendibility and the rank-into-rank cardinals.

mightwork, but my gut feeling (and I say this as a non-set-theorist) is that probably any one of these questions, say 1. to begin with, is going to be hard and would probably make a good MO question on its own. I'd actually be pleasantly surprised if someone at MO could answer such a question, as I would bet that among active MO members, you are currently the number 1 expert on these sorts of issues. $\endgroup$ – Todd Trimble♦ Feb 25 at 3:01