Is the topology that has the same sequential convergence with a metrizable topology equivalent as that topology?

No, this is not true in general. For instance, let $F$ be a nonprincipal ultrafilter on $X=\mathbb{N}$. Let $\mathscr{T}_1$ be the discrete topology, and let $\mathscr{T}_2$ consist of all sets that either are in $F$ or do not contain $0$. In both these topologies, a sequence converges iff it is eventually constant, and $\mathscr{T}_1$ is metrizable.

In the proof of Proposition IV.2.1, $\mathscr{T}_2$ is additionally known to be first-countable, since it is generated by countably many seminorms.

Inasmuch as a discrete space is metrizable and has no nontrivial convergent sequences, all you need for a counterexample is any nondiscrete space with no nontrivial convergent sequences. There are lots of examples; here are two that don't involve the axiom of choice.

Example 1. Let $X$ be an uncountable set, choose an element $p\in X,$ and let $$\tau=\{A\subseteq X:p\notin A\text{ or }|X\setminus A|\le\aleph_0\}.$$

Example 2. Let $X=\mathbb N=\{1,2,3,\dots\}$ and let $$\tau=\left\{A\subseteq\mathbb N:1\notin A\text{ or }\sum_{n\in\mathbb N\setminus A}\frac1n\lt\infty\right\}$$