In what sense of "structure" do group homomorphisms "preserve structure"? It is commonly said that group homomorphisms "preserve the structure of the group", e.g., from Wikipedia:


  The purpose of defining a group homomorphism as it is, is to create functions that preserve the algebraic structure.


Now, my default notion of structure is the one that holds that isomorphisms are structural identities---two isomorphic mathematical objects have the same structure. But that notion of structure requires preserving not just the operations on the elements of the domain but also requires the two sets of objects be equipollent.

But homomorphisms aren't, in general, reversible---or else they'd be isomorphisms. Since homomorphisms don't preserve size, in what sense does it preserve "structure"? Is there any more to the notion of structure here than the simple definition of algebraic structure as a set endowed with certain operations?

Toy example illustrating the problem: suppose you have a group homomorphism between groups $G$ and $H$ (but no isomorphism). Suppose also that you have an isomorphism between $G$ and some further group $J$. Since there is a homomorphism between $G$ and $H$, the structure in $G$ is preserved in $H$. Since there is an isomorphism between $G$ and $J$ they have the same structure. But presumably if $H$ preserves the structure of $G$ then they have the same structure. That can't be true, though, since they aren't isomorphic. It seems then that either "structure" means two different things when discussing isomorphisms and homomorphisms, or "preserves" doesn't mean that the homomorphic sets have the same structure (and then I have no idea what "preserves" actually means in this context).

I'm clearly confused about something. I'm just not sure what. 

TL;DR What is the difference between the notions of "structure" when you say that isomorphic structures have the same structure and when you say that a group homomorphism preserves the group structure?
The "structure"  preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 - c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, r\mapsto \bar r\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2  = (\bar b - \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$ For example,  a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all of its elements to see if they are roots, e.g. common parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence  $\ne 0;\,$ equivalently $\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence it has no integer roots.  This idea generalizes as follows

Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients 
has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd. 

Proof $\ $ If so then $\  \color{#0a0}{f(0)} \equiv  1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED 

In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because  homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".
Functors are the most general case of structure preserving function. A functor is a function that sends objects to objects and arrows to arrows in such a way that the structure is preserved which means 3 things: 
1)an identity arrow is sent into an identity arrow
2)given an arrow between two objects, the image of the arrow is another arrow between the images of the objects
3)operation of arrow is preserved (composition is the general case of an associative operation)

A group is a one object category, so that we really only have to pay attention to 1) and 3) which means that the function preserves unit and operation between elements. The last condition means that the function ¨commutes¨ with the operation in the sense that you get the same result from operating two elements and mapping the result, or if you map the original elements and then operate them in the image structure.

One thing that seems to confuse also is the fact that a structure preserving function does not have to bijective. For example, if the morphism $\phi:G\rightarrow H$ is not surjective this means we are mapping the group $G$ into a subgroup of $H$ (that is why the structure is preserved even if we dont cover all tof the image). When the morphism is not injective this means we dont preserve distinctness (there will be at least one pair of different elements in G mapped into the same element of H).