Can a Cauchy sequence converge for one metric while not converging for another?

Let $X=[0,\infty)$, let $d(x,y)=\lvert x-y\rvert$ and let$$e(x,y)=\begin{cases}\lvert x-y\rvert&\text{ if }x,y\neq0\\\lvert x+1\rvert&\text{ if }x\neq0\text{ and }y=0\\\lvert y+1\rvert&\text{ if }x=0\text{ and }y\neq0\\0&\text{ if }x,y=0.\end{cases}$$Then the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is a Cauchy sequence with respect to both metrics, but it converges only in $(X,d)$.

You can always have some artificial example where you just "move the limit elsewhere". For example, let $X=\mathbb{R_{\ge 0}}$, $d_1$ be the Euclidean metric and $d_2(x, y)=|\hat{x}-\hat{y}|$, where $ \hat{x}=-1$ if $x=0$ and $ \hat{x}=x$ otherwise. Then the sequence $x_n=\frac 1n$ converges in $(X, d_1)$ but not in $(X, d_2)$.

Depending on what you mean, the answer is "no". In particular, if by "two metrics on the one space" you mean two metrics on the same set that induce the same topology, then the answer is no because convergence is a topological property.

However the other answers are correct if the induced topologies are allowed to be different.