On HTT's remark 2.4.1.9

Question

Background
I have some difficulty to understand remark 2.4.1.9 in Lurie's HTT, which says a $p$-Cartesian edge is determined up to equivalence.
To be more precise, let $p:X\to S$ be an inner fibration of simplicial sets, let $x$ be a vertex of $X$, and let $\bar{f}:\bar{y}\to p(x)$ be an edge of $S$ ending at $p(x)$. If there are two $p$-Cartesian edges $f:y\to x$ and $f':y'\to x$ in $X$ with $p(f)=p(f')=\bar{f}$, then we have that $f$ and $f'$ are strong final objects of the $\infty$-category $K\stackrel{\text{def}}{=}X_{/x} \times_{S{/p(x)}}\{\bar{f}\}$, and thus they are equivalent $K$.
That's true. Since for a $p$-Cartesian edge $f$, $$q:X_{/f}\to X_{/x}\times_{S_{/p(x)}}S_{/p(f)}$$ is a trivial fibration, and $K_{/\{f\}}\to K$ is a pullback of $q$.

My Question:
But since Lurie's original words are "($p$-Cartesian edge) is therefore determined up to equivalence by $\bar{f}$ and $x$", I think we should conversely have that an object who is equivalent with a $p$-Cartesian edge is also a $p$-Cartesian edge. And he indeed utilized such a claim in the proof of Proposition 2.4.2.4 in HTT. However I cannot figure out a detailed proof. I hope someone can help me.

Answer 1

If you have a map $g$ in $X_{/x}$ from $f$ to $f'$, it is, by definition, a 2-simplex $\sigma$ of $X$, with faces $f'$, $f$, and $\bar g$, expressing that $f = f' \circ \bar g$.

If the map $g$ is an equivalence, then $\bar g$ is an equivalence (because $\bar g$ is the image of $g$ under the functor $X_{/x} \to X$). HTT 2.4.1.5 says that equivalences are $p$-Cartesian, so $\bar g$ is $p$-Cartesian. HTT 2.4.1.7 implies that composites of $p$-Cartesian edges are $p$-Cartesian, so if $f'$ is $p$-Cartesian, so is $f$. Applying the same to an inverse of $g$ gives the converse.

Answer 2

Take a look at Proposition 55.6 in Charles Rezk's notes Stuff about quasi-categories, (which is the formal meaning of Lurie's words). It says that fix an edge $\bar{f}:\bar{y}\to p(x)$ in $S$, the "space" of $p$-Cartesian lifts of $\bar{f}$ is contractible.

PS: I should have posted this as a comment, but I don't have enough reputation to do so...