Let $$A = \begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_d \\ a^2_1 & a^2_1 & \cdots & a^2_d \\ \vdots & \vdots & \ddots \ & \vdots \\ a^d_1 & a^d_2 & \cdots & a^d_d \end{pmatrix}$$ be a square matrix of dimension $d$ and rank $r$ which satisfies the homogeneous linear system
$$Ax = 0.$$
If we say that the rows of $A$ are such that not all entries of each are equal, that is; if $a_i$ is a row of A, then there is exists no constant $\alpha$ such that $\alpha \begin{pmatrix} 1&1& \cdots &1\end{pmatrix} = a_i$ for all $i$, is it true that the rank of $\tilde{A}$ is $r+1$ where
$$\tilde{A} = \begin{pmatrix}1&1& \cdots & 1\\ a^1_1 & a^1_2 & \cdots & a^1_d \\ a^2_1 & a^2_1 & \cdots & a^2_d \\ \vdots & \vdots & \ddots \ & \vdots \\ a^d_1 & a^d_2 & \cdots & a^d_d \end{pmatrix}?$$
I'm not sure how I would go about this other than just pure intuition. This condition I mention on the rows of $A$ would ensure that no row operation could completely eliminate any row of $\tilde{A}$ using this new row we introduced on the top that we couldn't already eliminate in $A$ before.
The answer to your question is no. As an example, consider the matrices $$ A = \pmatrix{1&1&0\\1&1&0\\0&0&1}, \quad \tilde A = \pmatrix{1&1&1\\1&1&0\\1&1&0\\0&0&1}. $$ $A$ has no constant rows, but $A$ and $\tilde A$ both have rank $2$.