Are two norms equivalent if they induce the same topology on a vector space?

Whenever two norms are equivalent in the sense that $\|x\|_1\le c_1\cdot \|x\|_2$ and $\|x\|_2\le c_2\cdot \|x\|_1$, they generate the same topology. Is the reverse also true, i.e. if a topology is generated by two different norms, are the norms equivalent in the above sense? We know this to be true for $\mathbb{R}^n$, but is it generally true, and if not, what are some counterexamples?

What about a Hausdorff, translationally invariant vector space topology generated by two different metrics, are the metrics equivalent in the sense $d_1(x,y)\le c_1\cdot d_2(x,y)$ and vice versa?

First, let's have a look at the case of a metric space $X$ and two metrics $d_1$ and $d_2$ on it. If these are equivalent, meaning $d_1(x,y) \leq c_1 d_2(x,y)$ and $d_2(x,y)\leq c_2 d_1(x,y)$ for some real numbers $c_1,c_2 > 0$ and all $x,y \in X$, we get the following inclusions for open balls $B_{r}^i(x) = \{y \in X \:|\: d_i(x,y) < r\}$: $$ B^2_{\frac{r}{c_1}}(x) \subseteq B^1_r(x) $$ $$ B^1_{\frac{r}{c_2}}(x) \subseteq B^2_r(x) $$ for all $x \in X$ and $r > 0$. As every open subset is a union of open balls, it follows that the metrics induce the same topology. The case for norms is a corollary now.

Conversely, if the topologies induced by two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $X$ are identical, we have an inclusion $B_r^1(0) \subset B^2_1(0)$ for some $r > 0$. Let $x \in X\setminus{\{0\}}$ and set $y = \frac{rx}{2\|x\|_1}$. Then $\|y\|_1 = \frac{r}{2} < r$, hence $\|y\|_2 < 1$ which shows $\|x\|_2 \leq \frac{2}{r} \|x\|_1$. By symmetry, you get the desired equivalence.

Now the bad news: Two metrics inducing the same topology do not have to be equivalent: Take $X = \mathbb{R}$ and $d_1$ the normal norm induced metric and set $d_2 = \frac{d_1}{d_1 + 1}$. It is not difficult to show that $d_1$ and $d_2$ induce the same topology, but there is no $c > 0$ with $ d_1(x,y) \leq c \cdot d_2(x,y)$ as the inequality $n \leq c \frac{n}{n + 1}$ does not hold for $n > c$.