If $\{x_1,x_2,\cdots,x_n\}$ is a basis, is $\{x_1+x_2,x_2+x_3,\cdots,x_n+x_1\}$ a basis too?

It is a matter to show whether the $n \times n$ tranformation matrix $$ \left( {\matrix{ 1 & 1 & 0 & \cdots & 0 \cr 0 & 1 & 1 & \cdots & 0 \cr 0 & 0 & 1 & \ddots & \vdots \cr \vdots & \vdots & \vdots & \ddots & 1 \cr 1 & 0 & 0 & \cdots & 1 \cr } } \right) $$ is invertible or not

By developing its determinant on the first column it is easy to show that it is $2$ if $n$ is odd and null if $n$ is even (for $3 \le n$).

Let there exist scalars $c_j$'s such that the following linear combination is equal to $0.$
$c_1(x_1+x_2)+c_2(x_2+x_3)+\ldots+c_{n-1}(x_{n-1}+x_n)+c_n(x_n+x_1)=0$
Case 1: $n$ is even:

Let $n=2m,\ m\in\Bbb N$. \begin{aligned}&c_1(x_1+x_2)+c_2(x_2+x_3)+\ldots+c_{2m-1}(x_{2m-1}+x_{2m})+c_{2m}(x_{2m}+x_1)=0\\\implies& (c_1+c_{2m})x_1+(c_1+c_2)x_2+(c_2+c_3)x_3+\ldots+(c_{2m-2}+c_{2m-1})x_{2m-1}+(c_{2m-1}+c_{2m})x_{2m}=0\end{aligned}

Hence,
$c_i+c_{i+1}=0, i=1,2,\ldots,2m-1$ and $c_{2m}+c_1=0\tag 1$
Now, note that \begin{aligned}c_{2i+1}&=c_1, i=1,2,\ldots,m-1&\\&&\text{and}&\\c_2&=c_{2i},i=2,3,\ldots,m.&\end{aligned} By $(1)$, $c_1+c_{2m}= c_1+c_2=0.$

Take $c_1= 2$, say, then,clearly $c_2=-2, c_3=2$ etc.
Hence, we don't necessarily have $c_i=0\ \forall i=1,2,3,\ldots,n$.

Thus, $x_1+x_2,x_2+x_3,\ldots,x_n+x_1$ are not linearly independent and thus can't be basis.

Case 2: $n$ is odd:

Let $n=2k+1, k\in\Bbb N$.
Proceed as in case $(1)$ above to get a system of linear equations similar to $(1)$
$c_i+c_{i+1}=0,i=1,2,\ldots,2k$ and $c_{2k+1}+c_1=0\tag{2}$ Again, note that \begin{aligned}c_{2i+1}&=c_1, i=1,2,\ldots,k&\\&&\text{and}\\c_2&=c_{2i},i=2,3,\ldots,k.&\end{aligned}

By $(2)$, \begin{aligned}c_1+c_{2k+1}= c_1+c_1&=0\\\implies c_1&=0=c_3=\ldots=c_{2k+1}\end{aligned}

and hence, again by $(2), c_2=c_4=\ldots=0$. Thus, in this case, $x_1+x_2,x_2+x_3,\ldots,x_n+x_1$ are linearly independent and thus form a basis.

To prove your claim where $n\geq 3$ is odd you have to show two properties:

1.) Let be $v\in V$ then $v$ must be contained in the linear span of $\{ x_1+x_2, x_2+x_3+, ..., x_n+x_1\}$. For sake of simplicity we define: $\{ x_1+x_2, x_2+x_3+, ..., x_n+x_1\}:=\{y_1, y_2, ...,y_n\}$.

2.) The vectors $y_1, y_2, ...,y_n$ must be linearly independent.

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Regarding 1.):

If $v\in V$ then there exists a linear combination of $v$ regarding the initial basis $x_1, x_2, ..., x_n$, namely $v=\sum\limits_{i=1}^n a_ix_i$. Then $v\in\{ y_1,y_2,..., y_n\}$ iff there exists a linear combination $v=\sum\limits_{i=1}^n b_i y_i$. If we simply set $b_j:= \sum\limits_{i=1}^j a_i (-1)^{j-i}+(-1)^j a_n$ where $j\in\{1,2,...,n\}$, then via induction it holds $v=\sum\limits_{i=1}^n b_i y_i=\sum\limits_{i=1}^n a_ix_i$. So the arbitrarily chosen vector $v$ is element of the span of $\{y_1, y_2, ...,y_n\}$.

Regarding 2.):

Let be $c_1,c_2,..., c_n$ some coefficients from the associated field, then we know that:

the vectors $y_1, y_2, ...,y_n$ are linearly independent iff $\sum\limits_{i=1}^n c_iy_i=0 \Rightarrow c_1=c_2=...=c_n=0$.

Let's assume $y_1, y_2, ...,y_n$ were not linearly independent then there must be at least one $i\in\{1, 2, ..., n\}$ with $c_i\neq 0$ such that:

$\sum\limits_{i=1}^n c_iy_i= 0$. By applying the inital assumption this leads to $0=\sum\limits_{i=1}^n c_iy_i= (c_1+c_n)x_1+ (c_1+c_2)x_2+ (c_2+c_3)x_3+ ...+c_n x_n$.

This is a contradiction because $x_1, x_2, ...x_n$ are linearly independent. So the vectors $y_1, y_2,...y_n$ must be linearly independent. Hence, the vectors $\{y_1, y_2, ...,y_n\}$ are another basis of $V$.