Let us suppose that the sequence $1, x_2, \dots, x_n, 1$ satisfy the condition of the problem. Then
$$|1 - x_2| = |x_2 - x_3| = \dots = |x_n - 1| = C. \quad (1)$$
Also
$$x_n - 1 = (x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + \dots + (x_2 - 1) \quad (2)$$
where the number of pairs of brackets of the right-hand side is $n-1$. We consider two cases: $n$ is odd and $n$ is even number.

Case 1. $n$ is odd, i.e. $n-1$ is even number.
Then the equality (2) get the form $\Delta_1 C = \underbrace{\Delta_2 C \Delta_3 C \dots \Delta_n C}_{n-1}$, where $\Delta_i \in \{-, +\}$, $1 \le i \le n$. The choice of $\Delta_i \in \{-, +\}$, $2 \le i \le n$ determine a sequence $1, x_2, \dots, x_n, 1$, satisfying the sequence (1). Let we

note that the sum of the right-hand side of the equality $\Delta_1 C = \underbrace{\Delta_2 C \Delta_3 C \dots \Delta_n C}_{n-1}$ (for one choice of $\Delta_i \in \{-, +\}, 2 \le i \le n$) is even number of times $C$, and the left-hand side is $\pm C$, i.e. odd number times $C$. Therefore, the right-hand side is equal to $\pm 2kC$, for some $k \in \mathbb{N}$, and the left-hand side is equal to $\pm C$. The last is possible only if $C=0$ and then we obtain the sequence $1=x_2=\dots=x_n$, i.e. there exists only one sequence with the desired property.

Case 2. $n$ is even, i.e. $n-1$ is odd number.
If $C=0$, then $1=x_2=\dots=x_n$. Let $C>0$. Then again from the equality (2) we obtain
$$\Delta_1 C = \underbrace{\Delta_2 C \Delta_3 C \dots \Delta_n C}_{n-1}.$$
Hence, the problem can be transformed to find the number of choices of $\Delta_i \in \{-, +\}, 2 \le i \le n$ such that $\Delta_1 C = \underbrace{\Delta_2 C \Delta_3 C \dots \Delta_n C}_{n-1}$. If on the left-hand side of the equality we have $+C$, then it must, on the right-hand side of the equality, the difference between the number of the positive and the number of the negative constants to be exactly 1, i.e. $+C$ must appear $\frac{n-2}{2}+1=\frac{n}{2}$ times, and $-C$ appears exactly $\frac{n-2}{2}$ ways. Such type of choices can be made on $\binom{n-1}{n/2} = \binom{n-1}{(n-2)/2}$ ways (for a sequence with length $n-1$ we choose $n/2$ positions where we will put $+C$, and on the rest we will put $-C$). On the exact same number of ways can be written the sum of the right-hand side is the sum of the left-hand side is $-C$. So, the total number of such representations is $2\binom{n-1}{n/2}$.
Finally, if with $r = r(n, C)$ we denote the number of the sequences satisfying the condition of the problem, then
$$r = \begin{cases} 1, & C=0 \\ 0, & n \text{ is odd and } C > 0. \\ 2\binom{n-1}{n/2}, & n \text{ is even and } C > 0. \end{cases}$$