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
Assumptions
The initial simplex tableau for your problem(, which is already in canoncial form,) will be
\begin{array}{cc|c} A & I & b \\ \hline -c^T & 0 & 0 \end{array}
We multiply the $m$ constraints by $B^{-1}$. Note that $\mathbf{e}_1,\dots,\mathbf{e}_n$ can be found in $[B^{-1}A\;B^{-1}]$ since $B$ is chosen from columns of $[A\;I]$. We add a linear combination of the first $m$ row to the last row.
\begin{array}{cc|c} B^{-1}A & B^{-1} & B^{-1}b \\ \hline c_B^T B^{-1}A-c^T & c_B^T B^{-1} & c_B^T B^{-1}b \tag{*} \label{t1} \end{array}
Why is $c_B^T$ chosen? We write $x_B$ as $(x_{B_1},\dots,x_{B_n})^T$. Choose an index $i$ in $\{1,\dots,n\}$, and suppose that $x_j = x_{B_i}$
Case 1: $1\le j \le n$. In the $j$-th column of the last row,
$$(c_B^T B^{-1}A-c^T)_j = c_B^T B^{-1} \mathbf{a}_j -c_j = c_B^T \mathbf{e}_i - c_j = c_{B_i} - c_j = 0.$$
Case 2: $n+1 \le j \le m+n$. Note that $c_{B_i}= 0$. In the $j$-th column of the last row,
$$(c_B^T B^{-1})_{j - n} = c_B^T \mathbf{e}_i = c_{B_i} = 0.$$
Therefore, the entries corresponding to the basic variables in the last row in tableau $\eqref{t1}$ will be zero. Then, $\eqref{t1}$ is a simplex tableau that you can compute from the given optimal solution.
Computational results for your example
Here's the GNU Octave code for finding the optimal tableau. The syntax should be the same in MATLAB except for the solver
glpk. You may test the code at Octave Online.The computer output
The optimal tableau:
\begin{equation*} \begin{array}{ccccccc|l} & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \\ \hline x_2 & \frac23 & 1 & \frac45 & 0 & \frac{1}{150} & 0 & 400 \\ s_1 & 16 & 0 & \frac{432}{5} & 1 & -\frac{32}{25} & 0 & 43200 \\ s_3 & \frac13 & 0 & \frac15 & 0 & -\frac{1}{150} & 1 & 100 \\ \hline & \frac{20}{3} & 0 & 0 & 0 & \frac23 & 0 & 40000 \\ \end{array} \end{equation*}