On non-isomorphic and isomorphic Steiner Triple System! in this wikipedia page, it says the following

Up to isomorphism, the STS(7) and STS(9) are unique, there are two STS(13)s, 80 STS(15)s, and 11,084,874,829 STS(19)s

Now, in wolframe page it says the following

The numbers of nonisomorphic Steiner triple systems S(v) of orders v=7, 9, 13, 15, 19, ... (i.e., 6k+1,3) are 1, 1, 2, 80, 11084874829, ...

it seems that "up to isomorphism" is the same as "nonisomorphism". According to what I know is that "isomorphic" means that we can map something to something else. But, "non-isomorphic" means that we cannot map something to something else. Now, in could you tell me what is different between "up to isomorphism" and "non-isomorphism"? Are they mean the same?
Suppose we have STS(7). Now, since number of non-isomorphic is 1 in STS(7), there is only one design in STS(7) that cannot be mapped to any other designs of STS(7) among all 6! of different permutation of different mapping, Is what I understand correct?!! In fact I would like to know how to show that STS(7) have only one isomorphism? Is it a theorem or just searching among all permutation of different mapping?!!
Thank you
 A: Counting things "up to isomorphism" means finding the number of "nonisomorphic" things. Let me take $S(7)$ as an example (and abbreviate $STS(v)$ to just $S(v)$, as in the Wolfram link).
Here's a sample $S(7)$ (rows are blocks in the design; so $\{1,2,4\}$ is in there, etc.):
\begin{array}{|c|c|c|c|c|c|c|}\hline
1 & 2 &   & 4 &   &   &   \\\hline
  & 2 & 3 &   & 5 &   &   \\\hline
  &   & 3 & 4 &   & 6 &   \\\hline
  &   &   & 4 & 5 &   & 7 \\\hline
1 &   &   &   & 5 & 6 &   \\\hline
  & 2 &   &   &   & 6 & 7 \\\hline
1 &   & 3 &   &   &   & 7 \\\hline
\end{array}
Here's another $S(7)$:
\begin{array}{|c|c|c|c|c|c|c|}\hline
  &   &   &   & 5 & 6 & 7 \\\hline
  &   & 3 & 4 &   &   & 7 \\\hline
1 & 2 &   &   &   &   & 7 \\\hline
  & 2 &   & 4 &   & 6 &   \\\hline
1 &   & 3 &   &   & 6 &   \\\hline
  & 2 & 3 &   & 5 &   &   \\\hline
1 &   &   & 4 & 5 &   &   \\\hline
\end{array}
Now, these aren't literally the same; $\{5, 6, 7\}$ is a block in the second, not the first. But, they are isomorphic. Let's define a map $\varphi \colon [7] \to [7]$, where $[7]$ is our shorthand for $\{1, 2, \ldots, 7\}$.
$$\varphi = 
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 1 & 5 & 7 & 4 & 6 & 3 \\
\end{pmatrix}$$
So $1 \mapsto 2,\, 2 \mapsto 1,\ 3 \mapsto 5$, etc. This map will take lines in the first $S(7)$ to lines in the second $S(7)$ (I found the map by just drawing the Fano planes; this is probably the easiest way to see it's an isomorphism!).
So here, our two $S(7)$'s are isomorphic.

Suppose we have STS(7). Now, since number of non-isomorphic is 1 in STS(7), there is only one design in STS(7) that cannot be mapped to any other designs of STS(7) among all 6! of different permutation of different mapping, Is what I understand correct?!! 

Not exactly -- actually, kind of the exact opposite $\ddot \smile$. To say that there's only 1 nonisomorphic design $S(7)$ means that if you have two $S(7)$'s, they must be isomorphic to each other (as in our example). So starting with any 1 $S(7)$ design $\mathcal{D}$, there's an isomorphism $\varphi \colon \mathcal{D} \to \mathcal{D}'$ to any other $S(7)$ design $\mathcal{D}'$.
Instead of saying "there are $n$ nonisomorphic $S(v)$ designs", it might be more clear to say that "there are $n$ isomorphism classes of $S(v)$ designs." This means that we can build a list of $n$ of those $S(v)$ designs such that


*

*None of them are isomorphic to anything else on that list; they're pairwise "nonisomorphic."

*Given any other $S(v)$ design, it must be isomorphic to something on our list of size $n$. 
In the $S(7)$ case, our list is just a single $S(7)$ design and any other $S(7)$ design must be isomorphic to ours.

In fact I would like to know how to show that STS(7) have only one isomorphism? Is it a theorem or just searching among all permutation of different mapping?!!

What you want to show is that if you have any two $S(7)$ designs, they must be isomorphic (or, that there's just one isomorphism class of $S(7)$ designs).
Here's a proof by Cameron (problem 3) that $S(7)$ is unique up to isomorphism, requiring essentially no theorems on designs or geometry. It basically constructs an isomorphism from any $S(7)$ to a fixed $S(7)$.
I'm sure that this result follows from theorems in various fields as well. People (not me, unfortunately -- maybe someone else will come along) know a lot about combinatorial designs, and I suspect there's a more high-brow proof that $S(7)$ is unique, up to isomorphism (meaning, all $S(7)$ designs are isomorphic to one another). 
