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).
