= Solution
For a <Lipschitz continuous> path $x:[0,T]\to V$, define its <signature of a bounded variation path> in the <truncated tensor algebra> by
$$
S_N(x)_{s,t}=\left(1,X^{(1)}_{s,t},\ldots,X^{(N)}_{s,t}\right),\qquad
X^{(k)}_{s,t}=\int_{s<u_1<\cdots<u_k<t}
dx_{u_1}\otimes\cdots\otimes dx_{u_k}.
$$
The <Riemann-Stieltjes integrals> exist since $x$ has <bounded variation>. Equivalently, since $x$ is absolutely continuous, the integrand is $x'(u_1)\otimes\cdots\otimes x'(u_k)$ with Lebesgue integration on the ordered simplex. The degree-zero component is $1$, and
$$
|X^{(k)}_{s,t}|\le\frac{|x|_{1\text{-var};[s,t]}^k}{k!}.
$$
The lift solves $dS=S\otimes dx$, starting at the identity. Since its velocities belong to the first layer, its endpoint lies in the <free step-N nilpotent Lie group>.
For $s\le t\le u$, partition the degree-$k$ integration simplex according to how many integration variables lie before $t$. If exactly $j$ do, the integral factors as the first $j$-fold integral on $[s,t]$ tensor the last $(k-j)$-fold integral on $[t,u]$. The boundaries with a time equal to $t$ have zero measure. Thus
$$
X^{(k)}_{s,u}=\sum_{j=0}^k X^{(j)}_{s,t}\otimes X^{(k-j)}_{t,u},
$$
which is precisely the <Chen identity>:
$$
\boxed{S_N(x)_{s,u}=S_N(x)_{s,t}\otimes S_N(x)_{t,u}.}
$$
This proves the identity at every level, including $k=0$, rather than just checking the first two levels.
For a strictly increasing continuously differentiable map $\psi:[a,b]\to[0,T]$, the actual <change of variables formula> is
$$
\boxed{S_N(x\circ\psi)_{a,b}=S_N(x)_{\psi(a),\psi(b)}.}
$$
To prove it, substitute $u_j=\psi(v_j)$ in each ordered integral. Strict increase preserves the order, and the almost-everywhere chain rule gives $d(x\circ\psi)(v)=x'(\psi(v))\psi'(v)\,dv$. The resulting integral is over $\psi(a)<u_1<\cdots<u_k<\psi(b)$. A vanishing derivative at some points causes no difficulty: monotone continuously differentiable substitution still applies, and the zero derivative contributes no variation.
Therefore \b[orientation-preserving reparametrisation of the whole segment preserves its <path signature>]. On $[0,1]$ the printed equality needs $\psi(0)=0$ and $\psi(1)=1$, equivalently that $\psi$ be onto. These endpoints are not forced by the stated assumptions. For example $x(t)=te_1$ and $\psi(t)=t/2$ satisfy those assumptions, but
$$
S_N(x)_{0,1}=\exp(e_1),\qquad
S_N(x\circ\psi)_{0,1}=\exp(e_1/2),
$$
which already differ at level one for every $N\ge1$. The endpoint-preserving interpretation proves the intended invariance, while the general formula states exactly what remains true without it.
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