Consider a continuous function $F(z,p)\colon \Omega\subset\mathbb{R}^N \times \mathbb{R}^N \to \mathbb{R}$ and the functional $$ \mathcal{F}(u)=\int_{a}^{b}{F(u(t),u'(t))\,dt}. $$ Prove that $\mathcal{F}$ is invariant under reparametrization (diffeomorphism) if and only if $F$ is homogenous respect the variable $p$, i.e. $F(z,\lambda p )=\lambda F(z,p)$ where $\lambda >0$. If $F$ is homogenous considering a regular reparametrization $t=\sigma(\tau)$ we obtain $$ \mathcal{F}(u)=\int_{a}^{b}{F(u(t),u'(t))\,dt}=\int_{\alpha}^{\beta}{F\Big(u(\sigma(\tau)),\frac{du}{dt}(\sigma(\tau))\Big)\sigma'(\tau)d\tau}. $$ where $\tau(\alpha)=a$ and $\tau(\beta)=b$. If $z(\tau)=u(\sigma(t))$ we obtain by the homogeneity condition $$ \mathcal{F}(u)=\int_{a}^{b}{F(u(t),u'(t))\,dt}=\int_{\alpha}^{\beta}{F(z(\tau)),z'(\tau))d\tau}= \mathcal{F}(z)=\mathcal{F}(u\circ \sigma). $$ I don't know how to procede with the other implication, i try starting from the condition $$ \int_{a}^{b}{F(u(t),u'(t))\,dt}=\int_{\alpha}^{\beta}{F(z(\tau)),z'(\tau))d\tau} $$ where $t=\sigma(\tau)$. Considering $\lambda(t)=\sigma'(\tau)\lvert_{\tau=\sigma^{-1}(t)}$, i obtain $$ \int_{\alpha}^{\beta}{F(z(\tau)),z'(\tau))d\tau}=\int_{a}^{b}{F(u(t),u'(t)\lambda(t))\frac{dt}{\lambda(t)}} $$ and consequently $$ \int_{a}^{b}{\bigg[F(u(t),u'(t)) - \frac{1}{\lambda(t)}F(u(t),u'(t)\lambda(t))\bigg]\,dt}=0. $$ Hence $F(u(t),u'(t)) - \frac{1}{\lambda(t)}F(u(t),u'(t)\lambda(t))=0$ for every $t \in [a,b]$ and for every $\lambda$ diffeomorphism, in particular for $\lambda(t)=c\in\mathbb{R}$.
It's correct o there is an other approach?