We call a set $S$ on the real line $R$ "superinvariant", if for any stretching $A$ of the set $S$ by the transformation taking $x$ to $A(x) = x_0 + a(x - x_0)$, where $a > 0$, there exists a transformation $B, B(x) = x + b$, such that the images of $S$ under $A$ and $B$ agree; i.e., for any $x \in S$, there is $y \in S$ such that $A(x) = B(y)$, and for any $t \in S$, there is a $u  \in S$ such that $B(t) = A(u).$ Determine all superinvariant sets.
1. **Define the problem and notation:**
   We call a set \( S \) on the real line \( \mathbb{R} \) "superinvariant" if for any stretching \( A \) of the set \( S \) by the transformation \( A(x) = x_0 + a(x - x_0) \), where \( a > 0 \), there exists a transformation \( B \), \( B(x) = x + b \), such that the images of \( S \) under \( A \) and \( B \) agree. This means for any \( x \in S \), there is \( y \in S \) such that \( A(x) = B(y) \), and for any \( t \in S \), there is a \( u \in S \) such that \( B(t) = A(u) \).

2. **Restate the problem in terms of functions:**
   Let \( p(a, b, c) \) be the proposition that \( A(S) = B(S) \), where \( A(x) = c + a(x - c) \) and \( B(x) = x + b \).

3. **Trivial case:**
   If \( a = 1 \), then choose \( b = 0 \), and we have \( p(a, b, c) \). So assume \( a \neq 1 \).

4. **Implications of \( p(a, b, c) \):**
   If \( x \in S \), then so is \( f(x) = B^{-1}A(x) \), and so is every integer power of \( f(x) \). Evaluating \( f(x) \) gives:
   \[
   f(x) = B^{-1}(A(x)) = B^{-1}(c + a(x - c)) = c + a(x - c) - b
   \]
   To simplify, let \( \gamma = c + \frac{b}{a - 1} \). Then:
   \[
   f(x) = \gamma + a(x - \gamma)
   \]
   This is a dilation with center \( \gamma \) and scale factor \( a \).

5. **Condition on the set \( S \):**
   For all valid \( p(a, b, c) \), we have \( x \in S \iff \gamma + a(x - \gamma) \in S \). This implies \( \gamma \) must be an accumulation point of \( S \).

6. **Choosing \( \gamma \) as a constant \( \Gamma \):**
   We can make \( \gamma \) equal to some constant \( \Gamma \) by always choosing \( b = (\Gamma - c)(a - 1) \). Now, let us draw the implications of \( p(a, (\Gamma - c)(a - 1), c) \) over all \( (a, c) \), where \( a > 0 \) and \( a \neq 1 \). This puts the following restrictions on \( S \):
   \[
   x \in S \iff \Gamma + a(x - \Gamma) \in S
   \]

7. **Possible sets \( S \):**
   This leads to a number of possible sets \( S \), based on answers to the following questions:
   - Does \( S \) contain an element less than \( \Gamma \)? Then \( S \) must contain all reals less than \( \Gamma \).
   - Does \( S \) contain an element more than \( \Gamma \)? Then \( S \) must contain all reals more than \( \Gamma \).
   - Does \( S \) contain \( \Gamma \)? Then \( S \) must contain \( \Gamma \) (obviously).

   Accordingly, there are 8 possible sets \( S \) that can result:
   \[
   S = \emptyset, \quad S = \{\Gamma\}, \quad S = (-\infty, \Gamma), \quad S = (-\infty, \Gamma], \quad S = (\Gamma, \infty), \quad S = [\Gamma, \infty), \quad S = (-\infty, \Gamma) \cup (\Gamma, \infty), \quad S = (-\infty, \infty)
   \]
   Any set of this form, for any \( \Gamma \), is superinvariant.

8. **Conclusion:**
   If the value of \( \gamma \) is not constant over all \( (a, b, c) \), it would imply \( S = (-\infty, \infty) \).

The final answer is \( \boxed{ S = \emptyset, \{\Gamma\}, (-\infty, \Gamma), (-\infty, \Gamma], (\Gamma, \infty), [\Gamma, \infty), (-\infty, \Gamma) \cup (\Gamma, \infty), (-\infty, \infty) } \).