Recall that $NFU$ is the Quine-Jensen system of set theory with a universal set; it is based on weakening the extensionality axiom of Quine's $NF$ so as to allow urelements.
Let $NFU^-$ be $NFU$ plus "every set is finite". As shown by Jensen (1969), $NFU^-$ is consistent relative to $PA$ (Peano arithmetic). $NFU^-$ provides a radically different "picture" of finite sets and numbers, since there is a universal set and therefore a last finite cardinal number in this theory.
The following summarizes our current knowedge of $NFU^-$.
1. [Solovay, unpublished]. $NFU^-$and $EFA$ (exponential function arithmetic) are equiconsistent. Moreover, this equiconsistency can be vertified in $SEFA$ (superexponential function arithmetic), but $EFA$ cannot verify that Con($EFA$) implies Con($NFU^-$). It can verify the other half of the equiconsistency.
2. [Joint result of Solovay and myself]. $PA$ is equiconsistent with the strengthening of $NFU^-$ obtained by adding the statement that expresses "every Cantorian set is strongly Cantorian". Again, this equiconsistency can be verified in $SEFA$, but not in $EFA$.
3. [My result]. There is a "natural" extension of $NFU^-$ that is equiconistent with second order arithmetic $\sf Z_2$.
For more detail and references, you can consult the following paper:
A. Enayat. From Bounded Arithmetic to Second Order Arithmetic via Automorphisms, in Logic in Tehran, Lecture Notes in Logic, vol. 26, Association for Symbolic Logic, 2006.
A preprint can be found here.
This question reminds me of a magical little-known theorem of Jean Pierre Ressayre that shows that every nonstandard model of $PA$ has a model of $ZF$ as a submodel of its Ackermann intepretation, more specifically:
Theorem. [Ressayre] Suppose $(M, +, \cdot)$ is a nonstandard model of $PA$, and $\in_{Ack}$ is the Ackermann epsilon on $M$, i.e., $a\in_{Ack}b$ iff $\mathcal{M}$ satisfies "the $a$-th digit in the binary expansion of $b$ is 1". Then for every consistent recursive extension $T$ of $ZF$ there is a subset $A$ of $M$ such that $(A,\in_{Ack})$ is a model of $T$.
Proof Outline: By Löwenheim-Skolem, it suffices to consider the case when $M$ is countable. Choose a nonstandard integer $k$ in $M$, and consider the submodel $M_k$ of $(M,\in_{Ack})$ consisting of sets of ordinal rank less than $k$ [as computed within $(M,\in_{Ack})$]. "Usual arguments" show that $(M,\in_{Ack})$ has a Tarskian truth-definition for $M_k$, which in turn implies that $(M_k,\in_{Ack})$ is recursively saturated. Since $M_k$ is also countable, $(M_k,\in_{Ack})$ must be resplendent [which means that it has an expansion to every recursive $\Sigma^1_1$ theory that its elementary diagram is consistent with].
Now add a new unary predicate symbol $A$ to the language ${\in}$ of set theory and consider the (recursive) theory $T^A$ consisting of sentences of the form $\phi^A$, where $\phi \in T$, and $\phi^A$ is obtained by relativizing every quantifier of $\phi$ to $A$. It is not hard to show that $T^A$ is consistent with the the elementary diagram of $(M_k,\in_{Ack})$, so by replendence the desired $A$ can be produced.
[I will be glad to add clarifications]
Ressayre's theorem appears in the following paper:
J. P. Ressayre, Introduction aux modèles récursivement saturés, Séminaire Général de Logique 1983–1984 (Paris, 1983–1984), 53–72, Publ. Math. Univ. Paris VII, 27, Univ. Paris VII, Paris, 1986.
Best Answer
Martin Gardner's Annotated Alice: The Definitive Edition says only this:
Gardner then goes on to say that the "Drawling-master" is a reference to the art critic John Ruskin, and gives a couple of paragraphs of biographical information about Ruskin. Given Gardner's extensive knowledge of Carrolliana, it seems likely that Carroll never published any further comments about the "different branches of Arithmetic."