Note that the "unstable operation" above won't give you a different solution. You've just replaced the basic variable $x_1 = 0$ by a nonbasic variable $x_4 = 0$, but the solution shown in the above optimal simplex tableau is still $(x_1,x_2,x_3,x_4,x_5,x_6)^T = (0,2,3,0,0,0)^T$.
Therefore, only "stable operations" with a non-zero variable $x_r \ne 0$ to be replaced by $x_j$ will give you a different basic optimal feasible solution.
It's easy to verify that any convex combination of a set of basic optimal feasible solution(s) is still an optimal feasible solution (since the feasible region in a linear program in convex), so the set of optimal feasible solution is convex (i.e. path connected). Hence, the answer to your question is yes.
(Added in response to the edit of the question)
You may just think of the graph so as to visualize your problem. A basic solution corresponds to an extreme point in the feasible region. (If $\mathbf x$ is an extreme point, then there doesn't exists two different $\mathbf u$ and $\mathbf v$ such that $\mathbf x$ is a convex combination of $\mathbf u$ and $\mathbf v$.) Form the graph of the set of all optimal feasible solutions of the linear program, and note its convexity (so it's a convex polygon), then eliminate its relative interior point. Take any two vertices of the remaining boundary points to be $X_1$ and $X_2$ in (1). Clearly, we can see a path running from $X_1$ to $X_2$ through adjacent vertices. (Assume that you have $m$ constraints and $n$ decision varibles, and $m < n$. Choose $n$ vectors in the $m$-by-$m+n$ matrix $A$ to form a basis matrix $B$. This is similar to choosing hyperplanes (representing the constraints) in $\Bbb R^{m+n}$ and find its intersection.) This corresponds to a (finite) series of steps of replacing $x_r$ by $x_j$, which is feasible. Since that's feasible, there exists a pivot operation in one of the simplex tableaux $T_1$ for $X_1$.
Let $y_{00} = \bar c$ be the objective function value, $y_{0j}$ be the $j$-th column of the objective function row (i.e. the row of $\bar c$), $y_{r0} = x_r$ be the $r$-th component of the current solution, and $y_{rj}$ be the $r,j$-th entry in the coefficient matrix.
\begin{matrix}
y_{0j}& y_{00}\\
y_{rj}& y_{r0}
\end{matrix}
In order not to change the value of $y_{00}$ after a pivot operation $y_{00}-\dfrac{y_{r0} y_{0j}}{y_{rj}}$, i.e. $$y_{00}-\dfrac{y_{r0} y_{0j}}{y_{rj}} = y_{00},$$ we have $y_{r0} y_{0j} = 0$. Since we want to move to another point, we must have $y_{r0} \ne 0$, which implies $y_{0j} = 0$, so the operation must be stable. Hence (1) is proved.
The proposed answer is not right.
The tableau has a reduced cost for a non-basic variable equal to $0$. This means that there are alternative optimal solutions. As it is not possible to go to another extreme point (the column $y_2$ is $\le 0$), there are optimal solutions which are obtained as:
$x^* + \lambda(-y_2,e_2)$, where $(-y_2,e_2)$ is an extreme direction, so the multiple solutions are:
$\left(\frac 23,0,\frac43\right)' + \lambda(1,1,0)' = \left(\frac23+\lambda, \lambda, \frac43\right)'$ for $\lambda \ge 0$.
So, as an example, you can observe that when $\lambda=1$ you obtain the solution $\left(\frac 53,1,\frac 43\right)$ which is feasible and also optimal.
Best Answer
From the given optimal solution $z(x_1,x_2,x_3,x_4,x_5)=z(30,0,0,0,10)=150$, you can find: $$x_1+5x_2+2x_3+x_4=b_1=\color{red}{30},\\ x_1-5x_2-6x_3+x_5=b_2=\color{red}{40}, $$ where $x_4$ and $x_5$ are the slack variables.
Since in the optimal table only the slack variable $x_4$ is replaced with $x_1$, then the intersection of $x_4$ row and $x_1$ column was the pivot element, which corresponds to the original value $1$. Hence the row remained from the initial table: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & & & & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & & & & 1 & 10 \\ \end{array}$$ In order to get $0$ as the first element in the row $z=5x_1+2x_2+3x_3$, where the first element was $-5$ initially, pivot row must have been multiplied by $5$ and added to it: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & & & & 1 & 10 \\ \end{array} $$ In order to get $0$ as the first element in the row $x_5$, where the first element was $1$ initially, pivot row must have been multiplied by $-1$ and added to it: $$\begin{array}{r|rrrrr|r} & x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\ x_5 & 0 & \color{red}{-10} & \color{red}{-8} & \color{red}{-1} & 1 & 10 \\ \end{array} $$ Note: I am not sure if this is ejecuting (executing). But it is reverse engineering (backwards working).