Remark: Some infinite sets may not be well-orderable at all, or even orderable. The statement "Every set is well-orderable." is one of the forms of the Axiom of Choice. The statement "Every set is orderable." is called the Ordering Principle, and is a weakened form of the Axiom of Choice. That said, you're quite correct that any infinite set that is (well-)orderable can be (well-)ordered in non-isomorphic ways.
So that we're on the same page, I'll be using the following
Definition: A (strict) total order is a structure $\langle X,R\rangle$--with $X$ a set and $R$ a relation--satisfying the following:
(i) $\forall x\in X$, we have $\neg(x\: R\: x)$. (irreflexivity)
(ii) $\forall x,y,z\in X$, we have $(x\: R\: y\wedge y\:R\: z)\Rightarrow(x\: R\: z)$. (transitivity)
(iii) $\forall x,y\in X$, we have $(x=y)\vee(x\:R\:y)\vee(y\:R\:x)$. (comparability)
UPDATE: I have replace the old claim (and its proof) with the following, more comprehensive, result.
Proposition: Let $\mathcal{A}$ be a set of axioms about structures $\langle X,R\rangle$--where $X$ is a set and $R$ is a relation--and let $\mathcal{A}'$ be the restriction of $\mathcal A$ to structures on finite sets (effectively, we're just adding an axiom of finiteness to $\mathcal A$ to get $\mathcal A'$). Suppose that every total order $\langle X,R\rangle$ satisfies the axioms of $\mathcal A$. Further suppose that there is some structure $\langle Y,S\rangle$ that is not a total order and which satisfies the axioms of $\mathcal A'$ (and so satisfies the axioms of $\mathcal A$). Then there is some finite $n$ such that $\mathcal A'$-structures on sets of cardinality $n$ need not be isomorphic.
Proof: Let $\langle Y,S\rangle$ be any $\mathcal A'$-structure that is not a total order, and put $n=|Y|$. Let $R$ be any total order relation on $Y$, so that $\langle Y,R\rangle$ is also a $\mathcal A'$-structure (being an $\mathcal A$-structure on a finite set). It suffices to show that $\langle Y,S\rangle$ is not isomorphic to $\langle Y,R\rangle$. By way of contradiction, suppose that they are isomorphic, and let $f$ be an isomorphism from $\langle Y,S\rangle$ to $\langle Y,R\rangle$. Since $\langle Y,S\rangle$ isn't a total order, then irreflexivity, transitivity, or comparability must fail.
If irreflexivity fails on $\langle Y,S\rangle$, then $\exists x\in Y$ such that $x\:S\:x$, but since $R$ is irreflexive on $Y$, then $\neg\bigl(f(x)\:R\:f(x)\bigr)$, which is impossible, since $f$ is an isomorphism. Thus, irreflexivity holds on $\langle Y,S\rangle.$
If transitivity fails on $\langle Y,S\rangle$, then $\exists x,y,z\in Y$ such that $(x\:S\:y)\wedge(y\:S\:z)\wedge\neg(x\:S\:z)$. Since $x\:S\:y$ and $f$ is an isomorphism, then $f(x)\:R\:f(y)$, and similarly, $f(y)\:R\:f(z)$, so by transitivity of $R$, we have $f(x)\:R\:f(z)$. But this is impossible, since $\neg(x\:S\:z)$ and $f$ is an isomorphism. Thus, transitivity holds on $\langle Y,S\rangle$.
Since irreflexivity and transitivity hold on $\langle Y,S\rangle$, then comparability fails--that is, there exist $x,y\in Y$ such that $(x\neq y)\wedge\neg(x\:S\:y)\wedge\neg(y\:S\:x)$. Since $f$ is an isomorphism, then $\bigl(f(x)\neq f(y)\bigr)\wedge\neg\bigl(f(x)\:R\:f(y)\bigr)\wedge\neg\bigl(f(y)\:R\:f(x)\bigr)$, so comparability fails on $\langle Y,R\rangle$. But $\langle Y,R\rangle$ is a total order, so this is impossible. Thus, we have our contradiction. $\Box$
Conclusion: It looks like the Proposition shows that your statement about total orders is the "simplest" of its kind, but it actually doesn't.
Consider the following axioms about structures $\langle X,R\rangle$:
(i$'$) If $X$ is finite, then $\forall x\in X$, we have $\neg(x\:R\:x)$.
(ii$'$) If $X$ is finite, then $\forall x,y,z\in X$, we have $(x\: R\: y\wedge y\:R\: z)\Rightarrow(x\: R\: z)$.
(iii$'$) If $X$ is finite, then $\forall x,y\in X$, we have $(x=y)\vee(x\:R\:y)\vee(y\:R\:x)$.
Clearly, every total order satisfies these axioms, and any structure on a finite set satisfying these axioms will be a total order. However, there are structures on infinite sets satisfying these axioms, but which aren't total orders--for example, given an infinite set $X$, define $R$ on $X$ by $x\:R\:y$ if and only if $x=y$. The result then crucially relies on the existence of a structure $\langle Y,S\rangle$ on a finite set $Y$ that satisfies the axioms of $\mathcal A$ and isn't a total order.
The axioms (i$'$), (ii$'$), (iii$'$) give us the "simplest" sort of structure that is uniquely determined on finite sets up to isomorphism. To my mind, though, it's rather contrived.