I'm trying to find the vertices of a simplex in the $5$-dimensional space, where the vertices are formed by only zeros and ones, similarly to these coordinates that represent a simplex in the $7$-dimensional space: $$\begin{align} \phantom{+}(0000000)+(0001111)+(0110011)+(0111100)\\ +(1010101)+(1011010)+(1100110)+(1101001) \end{align}$$
Is there an algorithm or method to calculate these kinds of simplexes?
This is not an answer but it's too long for a comment (EDIT - There is now an answer at the end). I don't think this is possible. I conjecture that there exists an $n$-simplex whose vertices have coordinates $(a_1,a_2,\ldots,a_n)$ where $a_i\in\{0,1\}$ if and only if $n=2^k-1$ for some positive integer $k$. I can prove the converse for now. (See my edit at the end. My conjecture is not true, but the proof below can be used to show that if an $n$-simplex with the required property exists, then a $(2n+1)$-simplex with the required property exists.)
Let $\oplus$ be the entry-wise addition modulo $2$. For example $$(0,1,1,0)\oplus(1,0,1,0)=(1,1,0,0).$$ Define $h(x)$ for $x\in\{0,1\}^n$ to be the number of ones in $x$. For example, $$h(0,0,1,1,1)=3.$$
For the converse, we prove inductively. Let $x_0^k,x_1^k,\ldots,x_{2^k-1}^k$ be the vertices for a given $k$ such that $x_0^k=(\underbrace{0,0,\ldots,0}_{2^k-1\ \text{zeros}})$, and each $x_i^k$ has exactly $2^{k-1}$ ones for $i=1,2,\ldots,2^k-1$.
We define the following vectors: $$X_i^k=\left(0,x_i^k\right)$$ and $$\bar{X}_i^k=(\underbrace{1,1,\ldots,1}_{2^k\ \text{ones}})\oplus X_i^k.$$ Now take $$x_{2i}^{k+1}=\left(x_i^k,X_i^k\right)$$ and $$x_{2i+1}^{k+1}=\left(x_i^k,\bar{X}_i^k\right).$$ We have \begin{align}h\left(x_{2i}^{k+1}\oplus x_{2i+1}^{k+1}\right)&=h\left(\underbrace{0,0,\ldots,0}_{2^k-1\ \text{zeros}},\underbrace{1,1,\ldots,1}_{2^k\ \text{ones}}\right)=2^k.\end{align} For $i\ne j$, \begin{align}h\left(x_{2i}^{k+1}\oplus x_{2j}^{k+1}\right)&=h\left(x_i^k\oplus x_j^k,0,x_i^k\oplus x_j^k\right)\\&=2\cdot h(x_i^k\oplus x_j^k)=2\cdot2^{k-1}=2^k,\end{align} \begin{align}h\left(x_{2i+1}^{k+1}\oplus x_{2j+1}^{k+1}\right)&=h\left(x_i^k\oplus x_j^k,0,x_i^k\oplus x_j^k\right)\\&=2\cdot h\left(x_i^k\oplus x_j^k\right)=2\cdot2^{k-1}=2^k,\end{align} and \begin{align}h\left(x_{2i}^{k+1}\oplus x_{2j+1}^{k+1}\right)&=h\left(x_i^k\oplus x_j^k,1,(\underbrace{1,1,\ldots,1}_{2^k-1\ \text{ones}})\oplus x_i^k\oplus x_j^k\right)\\&=h\left(x_i^k\oplus x_j^k\right)+2^k-h\left(x_i^k\oplus x_j^k\right)=2^k.\end{align}
For example, we start with $x_0^1=0$ and $x_1^1=1$. Then we have $$x_0^2=(0,0,0),x_1^2=(0,1,1),x_2^2=(1,0,1),x_3^2=(1,1,0).$$ Then $$x_0^3=(0,0,0,0,0,0,0), x_1^3=(0,0,0,1,1,1,1),$$ $$x_2^3=(0,1,1,0,0,1,1),x_3^3=(0,1,1,1,1,0,0),$$ $$x_4^3=(1,0,1,0,1,0,1),x_5^3=(1,0,1,1,0,1,0),$$ $$x_6^3=(1,1,0,0,1,1,0),x_7^3=(1,1,0,1,0,0,1).$$
As remarked by Empy2, if there exists Hadamard matrix $H_{n+1}$ of size $(n+1)\times (n+1)$, then there exists such an $n$-simplex. If $H_{n+1}$ exists, we can assume that its first row and its first column are both $e=(\underbrace{1, 1, \ldots, 1}_{n+1\ \text{ones}})\in\Bbb R^{n+1}$. Let $y_0,y_1,y_2,\ldots,y_{n}\in\Bbb \{1,-1\}^{n+1}$ be the columns of $H_{n+1}$. Since $y_0=e$ is orthogonal to $y_1,y_2,\ldots,y_n$, it follows that for $n>0$, $n$ must be odd and each $y_i$ has exactly $\frac{n+1}{2}$ ones for $i\ne 0$.
Define $$X_i=\frac{1}{2}\big(e-y_i\big).$$ Then the first position of $X_i$ is always $0$. We can take $x_0,x_1,\ldots,x_n\in\Bbb R^n$ to be $$X_i=(0,x_i).$$ Then $x_0=(0,0,\ldots,0)$ and for $i=1,2,\ldots,n$, $x_i$ has exactly $\frac{n+1}{2}$ ones.
If $i\ne 0$, we have $$h(x_0\oplus x_i)=h(x_i)=\frac{n+1}{2}.$$ If $i,j\ne 0$, we have $$h(x_i\oplus x_j)=h(x_i)+h(x_j)-2x_i\cdot x_j=(n+1)-\frac12(e-y_i)\cdot(e-y_j).$$ Thus for $i,j\ne 0$ and $i\ne j$, we get $$h(x_i\oplus x_j)=(n+1)-\frac{n+1}{2}=\frac{n+1}{2}.$$ This means $x_0,x_1,\ldots,x_n$ form an $n$-simplex.
I think the converse holds too. That is, I have the following conjecture. I guess that if $h(x_i)=\delta$ for $i\ne 0$, then we can assume wlog that $x_0,x_1,x_2,\ldots,x_{2\delta-1}$ live in $\mathbb{R}^{2\delta-1}$. The "wlog" assumption cannot yet be shown that it is indeed "wlog".
It is conjectured that $H_{n+1}$ exists iff $n=0$, $n=1$, or $n=4k-1$ for some positive integer $k$. (The forward implication is known and not difficult to prove. That is, if $H_{n+1}$ exists, then $n=0$, $n=1$, or $n=4k-1$. See page 6 here.) The conjecture is true at least for $n\le 100$. Therefore at least for $n\le 100$ and $n=4k-1$, you can find your $n$-simplex.
For the situation at hand ($n=5$), such a simplex doesn't exist. Suppose for contradiction that $x_0,x_1,x_2,x_3,x_4,x_5\in\{0,1\}^5$ exist so that $h(x_i\oplus x_j)=\delta$ for some fixed integer $\delta$. WLOG, $x_0=(0,0,0,0,0)$.
Note that $h(x_i)=h(x_i\oplus x_0)=\delta$ for $i\ne 0$. Therefore, $$\delta=h(x_i\oplus x_j)=h(x_i)+h(x_j)-2x_i\cdot x_j=2\delta-2x_i\cdot x_j.$$ This means $$\delta=2x_i\cdot x_j.$$ Hence $\delta$ is even. Write $\delta=2d$. Clearly, there are only two possible cases: $d=1$ or $d=2$.
If $d=2$, then $\delta=4$. There are exactly $5$ vectors in $\{0,1\}^5$ with $4$ ones: $$(0,1,1,1,1),(1,0,1,1,1),(1,1,0,1,1),(1,1,1,0,1),(1,1,1,1,0).$$ This means $$h(x_i\oplus x_j)=2\ne\delta$$ for $i,j\ne 0$ and $i\ne j$. So $d\ne 2$.
If $d=1$, then $\delta=2$. Wlog, we can take $$x_1=(0,0,0,1,1)\wedge x_2=(0,0,1,0,1).$$ Now, it follows that there is only one choice left for $x_3,x_4,x_5$, which is $(0,0,1,1,0)$. We have a contradiction again.