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 $n > c$.
