Let $(X,\mathcal T)$ and $(Y,\mathcal T')$ two homeomorphic topological spaces.
Q1) If $\mathcal T$ is metrizable, will $\mathcal T'$ be also metrizable ?
Q2) If yes, does both metric will be comparable ? In the sense : there are $A$ and $B$ s.t. $$ Ad_X(x,y)\leq d_Y(f(x),f(y))\leq Bd_X(x,y),\tag{E}$$
where $f:X\to Y$ is an homeomorphism.
Q3) If $Y=X$, and $(X,\mathcal T)$ and $(X,\mathcal T')$ are homeomorphic and $\mathcal T$ is metrizable, then does the two metric will be comparable in the following sense ? There is $A$ and $B$ s.t. $$Ad_X(x,y)\leq d_Y(x,y)\leq Bd_X(x,y) \ \ ?\tag{F}$$
Attempts
1) I think it is since "homeomorphic" mean that there are the same. But I can't prove it. In fact, I know how to prove that a space is not metrizable, but I always have problem to prove that it is metrizable.
2) I really would say no, since $d_Y(f(x),f(y))\leq C d_X(x,y)$ would mean that $f$ is Lipschitz, and a priori, there is no reason for this. But maybe in some case it can happen, and if $f$ is Lipschitz, would $E$ hold ? I wouldn't be surprise that $f^{-1}$ would also be Lipschitz, and thus $(E)$ indeed hold. But in fact, I really have difficulty to prove it. May be it's not true ?
3) I know this work for normed vector spaces in finite dimension... as many property, this could fail in infinite dimension. If it indeed fail, would $(E)$ work in this case ? i.e. if $f:(X,\mathcal T)\to (X,\mathcal T')$ is a homeomorphism, then does $$Ad_X(x,y)\leq d_X(f(x),f(y))\leq Bd_X(x,y) \ \ ?$$
Q1). As amsmath said, you can explicitly define the metric.
Q2 and Q3). No. Note that any strictly increasing, continuous $f: \mathbb{R} \to \mathbb{R}$ will be a homeomorphism, and you can of course find such an $f$ with $\inf_{x < y} \frac{f(y)-f(x)}{y-x} = 0$ and $\sup_{x < y} \frac{f(y)-f(x)}{y-x} = +\infty$.