If $\sup_{t\in[0,\:\tau]}\left\|T_t(x)-T_t(y)\right\|\le c\left\|x-y\right\|$, is $\frac{\partial T}{\partial t}$ is Lipschitz continuous as well?

Question

Let $\tau>0$, $d\in\mathbb N$ and $T:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ with $$\sup_{t\in[0,\:\tau]}\left\|T_t(x)-T_t(y)\right\|\le c\left\|x-y\right\|\;\;\;\text{for all }x,y\in\mathbb R^d\tag1$$ for some $c>0$. If $T$ is differentiable in the first argument, are we able to show that $\frac{\partial T}{\partial t}$ is Lipschitz continuous as well?

We may clearly write $$\frac{\partial T}{\partial t}(t,x)-\frac{\partial T}{\partial t}(t,y)=\lim_{h\to0}\frac{\left(T_{t+h}(x)-T_{t+h}(y)\right)-\left(T_t(x)-T_t(y)\right)}h\tag2$$ for all $t\in[0,\tau]$ and $x,y\in\mathbb R^d$.

Answer 1

$T_t(x)=t^{3/2} Tx$ where $T: \mathbb R^{d} \to \mathbb R^{d}$ is any non-zero linear map is a counter-example.