Let $ n\ge 3$ be a natural number. Find all nonconstant polynomials with real coeficcietns $ f_{1}\left(x\right),f_{2}\left(x\right),\ldots,f_{n}\left(x\right)$, for which
\[ f_{k}\left(x\right)f_{k+ 1}\left(x\right) = f_{k +1}\left(f_{k + 2}\left(x\right)\right), \quad  1\le k\le n,\]
for every real $ x$ (with $ f_{n +1}\left(x\right)\equiv f_{1}\left(x\right)$ and $ f_{n + 2}\left(x\right)\equiv f_{2}\left(x\right)$).
1. Let \( n \ge 3 \) be a natural number. We need to find all nonconstant polynomials with real coefficients \( f_1(x), f_2(x), \ldots, f_n(x) \) such that 
   \[
   f_k(x) f_{k+1}(x) = f_{k+1}(f_{k+2}(x)), \quad 1 \le k \le n,
   \]
   for every real \( x \), with \( f_{n+1}(x) \equiv f_1(x) \) and \( f_{n+2}(x) \equiv f_2(x) \).

2. Let \( d_k \) be the degree of \( f_k(x) \). The given equation implies:
   \[
   d_k + d_{k+1} = d_{k+1} d_{k+2}.
   \]
   Rearranging, we get:
   \[
   d_k = d_{k+1} (d_{k+2} - 1).
   \]

3. If \( d_{k+2} = 1 \), then \( d_k = 0 \), which is impossible since \( f_k \) is nonconstant. Therefore, \( d_{k+2} > 1 \) and hence \( d_k \geq d_{k+1} \).

4. The cyclic nature of the indices implies that \( d_k \) must be constant for all \( k \). Let \( d_k = d \) for all \( k \). Then:
   \[
   d = d (d - 1).
   \]
   Solving this, we get:
   \[
   d^2 - 2d = 0 \implies d(d - 2) = 0.
   \]
   Since \( d \neq 0 \), we have \( d = 2 \).

5. Therefore, each \( f_k(x) \) is a quadratic polynomial. Let:
   \[
   f_k(x) = a_k x^2 + b_k x + c_k.
   \]
   Substituting into the original equation, we get:
   \[
   (a_k x^2 + b_k x + c_k)(a_{k+1} x^2 + b_{k+1} x + c_{k+1}) = a_{k+1} (a_{k+2} x^2 + b_{k+2} x + c_{k+2})^2 + b_{k+1} (a_{k+2} x^2 + b_{k+2} x + c_{k+2}) + c_{k+1}.
   \]

6. Equating the coefficients of \( x^4 \) on both sides, we get:
   \[
   a_k a_{k+1} = a_{k+1} a_{k+2}^2.
   \]
   Since \( a_{k+1} \neq 0 \), we have:
   \[
   a_k = a_{k+2}^2.
   \]
   This implies:
   \[
   a_k = a_k^{2^n}.
   \]
   Therefore, \( a_k = 1 \) for all \( k \).

7. Now, the polynomials simplify to:
   \[
   f_k(x) = x^2 + b_k x + c_k.
   \]
   The equation becomes:
   \[
   (x^2 + b_k x + c_k)(x^2 + b_{k+1} x + c_{k+1}) = (x^2 + b_{k+2} x + c_{k+2})^2 + b_{k+1} (x^2 + b_{k+2} x + c_{k+2}) + c_{k+1}.
   \]

8. Equating the coefficients of \( x^3 \), we get:
   \[
   b_k + b_{k+1} = 2 b_{k+2}.
   \]
   This implies \( b_k = b \) for all \( k \), otherwise \( b_{n+1} \neq b_1 \).

9. Therefore, \( b_k = b \) for all \( k \). The equation now becomes:
   \[
   (x^2 + b x + c_k)(x^2 + b x + c_{k+1}) = (x^2 + b x + c_{k+2})^2 + b (x^2 + b x + c_{k+2}) + c_{k+1}.
   \]

10. Equating the coefficients of \( x^2 \), we get:
    \[
    c_k + c_{k+1} + b^2 = b^2 + 2 c_{k+2} + b.
    \]
    Simplifying, we get:
    \[
    c_k + c_{k+1} = 2 c_{k+2} + b.
    \]

11. Adding these \( n \) equalities, we get:
    \[
    \sum_{k=1}^n (c_k + c_{k+1}) = \sum_{k=1}^n (2 c_{k+2} + b).
    \]
    This simplifies to:
    \[
    2 \sum_{k=1}^n c_k = 2 \sum_{k=1}^n c_k + n b.
    \]
    Therefore, \( b = 0 \).

12. Hence, \( c_k = c \) for all \( k \). The equation now becomes:
    \[
    (x^2 + c)(x^2 + c) = (x^2 + c)^2 + c.
    \]
    Simplifying, we get:
    \[
    x^4 + 2 c x^2 + c^2 = x^4 + 2 c x^2 + c^2 + c.
    \]
    Therefore, \( c = 0 \).

13. Thus, the only solution is:
    \[
    f_k(x) = x^2.
    \]

The final answer is \( \boxed{ f_k(x) = x^2 } \).