Determine all triples $(x,y,z)$ of real numbers  that satisfy all of the following three equations:
$$\begin{cases}  \lfloor x \rfloor + \{y\} =z \\ \lfloor y \rfloor + \{z\} =x \\ \lfloor z \rfloor + \{x\} =y  \end{cases}$$
We are given the system of equations:
\[
\begin{cases}
\lfloor x \rfloor + \{y\} = z \\
\lfloor y \rfloor + \{z\} = x \\
\lfloor z \rfloor + \{x\} = y
\end{cases}
\]
where $\lfloor \cdot \rfloor$ denotes the floor function and $\{\cdot\}$ denotes the fractional part.

1. **Express the equations in terms of integer and fractional parts:**
   Let $x = \lfloor x \rfloor + \{x\}$, $y = \lfloor y \rfloor + \{y\}$, and $z = \lfloor z \rfloor + \{z\}$. Denote $\lfloor x \rfloor = a$, $\lfloor y \rfloor = b$, and $\lfloor z \rfloor = c$, where $a, b, c \in \mathbb{Z}$, and $\{x\} = u$, $\{y\} = v$, and $\{z\} = w$, where $0 \leq u, v, w < 1$.

2. **Rewrite the system using these notations:**
   \[
   \begin{cases}
   a + v = c + w \\
   b + w = a + u \\
   c + u = b + v
   \end{cases}
   \]

3. **Analyze the equations:**
   From the first equation, we have:
   \[
   a + v = c + w \implies v - w = c - a
   \]
   From the second equation, we have:
   \[
   b + w = a + u \implies w - u = a - b
   \]
   From the third equation, we have:
   \[
   c + u = b + v \implies u - v = b - c
   \]

4. **Combine the equations:**
   Adding the three derived equations:
   \[
   (v - w) + (w - u) + (u - v) = (c - a) + (a - b) + (b - c)
   \]
   Simplifies to:
   \[
   0 = 0
   \]
   This is always true, so it does not provide new information.

5. **Consider the implications of the derived equations:**
   \[
   v - w = c - a \\
   w - u = a - b \\
   u - v = b - c
   \]
   Adding these equations again, we get:
   \[
   (v - w) + (w - u) + (u - v) = 0 \implies 0 = 0
   \]
   This confirms that the system is consistent.

6. **Assume $x = y = z$:**
   If $x = y = z$, then $a = b = c$ and $u = v = w$. Substituting $x = y = z$ into the original equations:
   \[
   \lfloor x \rfloor + \{x\} = x \\
   \lfloor y \rfloor + \{y\} = y \\
   \lfloor z \rfloor + \{z\} = z
   \]
   These are trivially true since $\lfloor x \rfloor + \{x\} = x$ by definition.

Therefore, the only solution is when $x = y = z$.

The final answer is $\boxed{(x, y, z) \text{ such that } x = y = z}$.