Consider a real inner product space $V$, and a linearly independent vector sequence $v=\{v_k\}_{1\leq k\leq n}$. The Gram-Schmidt algorithm uses $v$ to produce an orthonormal sequence $e=\{e_k\}_{1\leq k\leq n}$ recursively, like so.
$$u_k = v_k - \sum_{j=1}^{k-1}\operatorname{proj}_{u_j}(v_k) \hspace{24mm} e_k = \frac{u_k}{||u_k||}$$
It can be proven inductively that each $k$ satisfies:
$$\operatorname{Span}\{v_1,\cdots,v_k\}=\operatorname{Span}\{e_1,\cdots, e_k\}$$
Using the properties of span, likewise:
$$\operatorname{Span}\{e_1,\cdots, e_k\}=\operatorname{Span}\{u_1,\cdots,u_k\}$$
Label $E_k=\operatorname{Span}\{v_1,\cdots,v_k\}$, then we have the following equalities.
$$E_k=\operatorname{Span}\{v_1,\cdots,v_k\} = \operatorname{Span}\{e_1,\cdots, e_k\} = \operatorname{Span}\{u_1,\cdots,u_k\}$$
Now, consider a second linearly independent vector sequence $v'=\{v'_k\}_{1\leq k\leq n}$, and define the $e'$ and $E'$ sequences analogously. We want to determine the necessary and sufficient conditions to have $e=e'$. By the previous observations, if $e=e'$ it is necessary to have $E=E'$. Using the contrapositive, this is enough to answer your second question.
If we have $v$ and $v'$ identical except some $i<k$ have $v_i,v_k$ swapped, then $e\neq e'$. Indeed, we have $v_i\notin \operatorname{Span}\{v_1,\cdots,v_{i-1},v_k\}=\operatorname{Span}\{v'_1,\cdots,v'_{i-1},v'_i\}=E'_i$, by linear independence, thus we have $E_i\neq E'_i$ so that $e_i\neq e'_i$, proving $e\neq e'$.
As pointed out by user ancient mathematician, each $e_k$ must be in the orthogonal complement of $E_{k-1}$, so that $e_k\in E_k\cap (E_{k-1})^\perp$. This space is one dimensional however, and thus has exactly two unit vectors, one of which is $e_k$, so the other is $-e_k$. It follows that any unit vector $u\in E_k\cap (E_{k-1})^\perp$ must have $u=\pm e_k$. In particular if we have $E=E'$, the following holds.
$$e_k\in E_k\cap (E_{k-1})^\perp = E'_k\cap (E'_{k-1})^\perp \ni e'_k$$
$$\implies e'_k=\pm e_k$$
So if $E=E'$, then all $k$ have $e'_k=\pm e_k$. Since $E_k$ is simply the span of some $e_j$, it is easily shown that, conversely, if all $k$ have $e'_k=\pm e_k$ then necessarily $E=E'$. Thus we have shown that all $k$ have $\operatorname{Span}\{v_1,\cdots,v_k\}=\operatorname{Span}\{v'_1,\cdots,v'_k\}$ if and only if all $k$ have $e'_k=\pm e_k$.
Moreover when $E=E'$, each $\left<e_k, e'_k\right> = \pm 1$, and since $e_k=e'_k$ if and only if $\left<e_k, e'_k\right> = 1$, then $e_k=e'_k$ if and only if $\left<e_k, e'_k\right> > 0$. Since $e'_k=\frac{u'_k}{||u'_k||}$, likewise $e_k=e'_k$ if and only if $\left<e_k, u'_k\right> > 0$. Since we define $u'_k=v'_k-\operatorname{proj}_{E_{k-1}}(v'_k)$, where the projection is necessarily perpendicular to $e_k$, then $\left<e_k, u'_k\right>=\left<e_k, v'_k\right>$, therefore $e_k=e'_k$ if and only if $\left<e_k, v'_k\right> > 0$. In geometric terms, this means the angle between $e_k$ and $v'_k$ is smaller than $90$ degrees.
Using the formula for distance $d(p_1,p_2)=||p_1-p_2||$, and the formula for norm $||p||=\sqrt{\left<p,p\right>}$, it is not too hard to show that $\left<e_k, v'_k\right> > 0$ if and only if $d(v'_k,e_k)<d(v'_k,-e_k)$. In other words, provided that $E=E'$, then $e'_k=e_k$ if and only if $v'_k$ is closer to $e_k$ than to its negation $-e_k$. This is enough to answer your first question.
We have $e=e'$ if and only if all $k$ have $\operatorname{Span}\{v_1,\cdots,v_k\}=\operatorname{Span}\{v'_1,\cdots,v'_k\}$, and we have the distance inequality $d(e_k,v'_k) < d(-e_k,v'_k)$.
So, the Gram-Schmidt algorithm will produce the same orthonormal basis if and only if each of these $E_k$ subspaces are identical to $E'_k$, and each $v'_k$ is "close enough" to $e_k$ that $-e_k$ is further away.
For some intuition, if you fix the sequence $v$ as constant, in general you'll find that for each fixed $k$, exactly half of all $v'$ sequences which satisfy $E=E'$ will also satisfy $e'_k=e_k$. Here I mean "half" in the sense that there will be some hyperplane that partitions the space, where if $v'_k$ is on one side then $e'_k=e_k$, and on the other side $e'_k=-e_k$.
