How can I find explicit examples of maximal orders of quaternion algebras that are not isomorphic? I know that there exist algorithms that will construct maximal orders of a quaternion algebra over, say, $\mathbb{Q}$. However, the implemented algorithms that I know of require that you input an order that is not necessarily maximal, which the algorithm then completes. Unfortunately, this does not help if you want examples of orders that are not isomorphic to one another.
The more examples (at least over $\mathbb{Q}$, but preferably over other number fields) that I can get, the better, but I would be happy even with a table of some known examples.
 A: I'm really only familiar with how to do this in Magma, so all of the commands I'll refer to below are going to be for Magma. If you don't have a Magma license then you can run the below commands for free on the Magma online calculator at: http://magma.maths.usyd.edu.au/calc/
Suppose that you want to list representatives for all of the isomorphism classes of maximal orders in a quaternion algebra over some number field. For the sake of concreteness let's say that your number field is $k=\mathbb Q(\sqrt{-10})$ and that the quaternion algebra you are interested in is the one with Hilbert symbol $A=\left(\frac{-1,-3}{k}\right)$.
$\texttt{> k<t>:=QuadraticField(-10);}$
$\texttt{> A<i,j,ij>:=QuaternionAlgebra<k|-1,-3>;}$
$\texttt{> C:=ConjugacyClasses(MaximalOrder(A));}$
$\texttt{> #C;}$
$\texttt{2}$
$\texttt{> IsConjugate(C[1],C[2]);}$
$\texttt{false}$
$\texttt{> Generators(C[1]);}$
$\texttt{[ 1, i, 1/2*i + 1/2*j, 1/2 + 1/2*t*i + 1/6*t*j + 1/6*ij ]}$
$\texttt{> Generators(C[2]);}$
$\texttt{[ 1, 2*i, 3*t*i, 1 + 1/2*i + 1/2*j, 1/2*(t + 2) + 1/4*(t + 2)*i + }$
$\texttt{1/4*(t + 2)*j, 1/2*(t + 1) + 1/4*(t + 4)*i - 1/12*t*j + 1/6*ij ]}$
As you can see, the above code constructs a set of representatives for the two isomorphism classes of maximal orders in $A$. The key command you'll need is the $\texttt{ConjugacyClasses()}$ command in Magma. You can read more about it here: http://magma.maths.usyd.edu.au/magma/handbook/text/1002#11397
Finally, you should keep in mind that over many number fields, all maximal orders in a given (indefinite) quaternion algebra are going to be conjugate to each other. This is the case for the number field $\mathbb Q$ for instance. The underlying issue is that if $k$ is a number field and $A$ is an indefinite quaternion algebra over $k$ then there is a finite extension $k_A$ of $k$ such that the type number of $A$ (i.e., the number of conjugacy classes of maximal orders of $A$) is equal to the degree $[k_A:k]$. Moreover, $k_A$ can be characterized as the maximal abelian extension of $k$ which has $2$-elementary Galois group, is unramified outside of the real places in $\mathrm{Ram}(A)$ and in which all finite primes of $\mathrm{Ram}(A)$ split completely. Here $\mathrm{Ram}(A)$ is the set of places of $k$ which ramify in $A$. In particular this means that $k_A$ is contained in the narrow class field of $k$. This implies that if $A$ is an indefinite quaternion algebra defined over a number field of narrow class number one then all maximal orders of $A$ are conjugate. For more details about this and some useful references you can check out, see Section 5.4 and Chapter 9 of these notes of mine (and the cited references).
A: Ben explained what happens for indefinite quaternion algebras, and in particular that you don't get such examples if your base field has narrow class number one.  Let me discuss the (totally) definite quaternion algebra case, where you get lots of such examples even over $F=\mathbb Q$.  I can tell you at least 2 methods.  The first you can easily do in Magma.  The second, you can do in Magma or Sage (or by hand if you are patient).  Possibly you can do these things in Pari/GP as well.
Method 1:
Let $F$ be a totally real number field and $B/F$ a totally definite quaternion algebra.  Let $O$ be any maximal order, and $Cl(O)$ the set of left ideal classes of $O$.  Let $I_1, ..., I_h$ be a set of representatives for $Cl(O)$.  
Theorem: Every type of maximal order (i.e., conjugacy class of maximal order) of $B$ is represented by some $O_r(I_i)$, the right order of $I_i$.
(The converse is not true, different ideal classes may have the same type of right order, but at least we have the type number is at most the class number $h$.)
Here's a simple example in Magma (unfortunately, I don't think the calculation of $Cl(O)$ is implemented in Sage) for $B$ the quaternion algebra of discriminant 11 over $F = \mathbb Q$.  Here $h=2$.
> QQ := RationalField();
> B := QuaternionAlgebra(QQ, -1,-11);
> O := MaximalOrder(B);                                                    
> Cl := LeftIdealClasses(O);
> O1 := RightOrder(Cl[1]);
> O2 := RightOrder(Cl[2]);
> IsIsomorphic(O1,O2);
false
> Basis(O1);
[ 1, i, 1/2*i + 1/2*k, 1/2 + 1/2*j ]
> Basis(O2);
[ 1/2 + 1/2*j + k, 1/4*i + 1/2*j + 5/4*k, j, 2*k ]

Method 2:
Your comment about not an algorithm for extending orders to maximal orders not being helpful is incorrect.  For instance, let's work with definite $B/\mathbb Q$.  Take some imaginary quadratic $E$ embedding in $B$.  Then the ring of integers $O_E$ of $E$ will not in general embed in all maximal orders (if $E$ has small class number).  So as you vary $O_E$ and construct maximal orders containing $O_E$ you should often find non-isomorphic ones.
My recent student, Jordan Wiebe, used a similar idea to construct bases of orders of level $N$ in quaternion algebras $B$:
https://arxiv.org/abs/1810.05249
In our convention of level $N=D$, the discriminant, corresponds to maximal orders.  This basis depends on choosing several parameters (though things are much simpler in the maximal order case), one of which is a quadratic order $o \subset E$ that will be contained in your quaternionic order.  So you can also use this to construct various maximal orders, and test them for isomorphism.  Currently you can get my student's Sage code from his webpage, though this might disappear at some point since he has graduated.
I don't remember, but I imagine Sage has code to test whether different orders in the same quaternion algebra are isomorphic.  FYI here is an elementary method you can use that differentiates most orders: if you have two orders $O_1$ and $O_2$, they can only be isomorphic if they have the same theta series, i.e., if they have the same number of elements of norm $n$ for every $n$.  So you can just count the number of such elements using lattice/quadratic form methods for small $n$, and if you find a discrepancy then $O_1 \not \simeq O_2$.  (Note it is quite rare, but an interesting problem to find, non-isomorphic lattices with the same theta series.)
