In most (but not all) cases, this relation is just joint similarity. This fact relies on the following proposition.
Proposition: Let $K$ be a field. Then the only automorphisms of the $K$-algebra $M_d(K)$ are inner automorphisms. In other words, if $\phi:M_d(K)\rightarrow M_d(K)$ is a $K$-algebra automorphism, then there is some invertible $X$ with $\phi(Y)=XYX^{-1}$ for all $Y\in M_d(K)$.
Here is a proof of the above proposition.
Suppose that $(A_1,\dots,A_r),(B_1,\dots,B_r)$ are $d\times d$-matrices over a field $K$ where $(A_1,\dots,A_r),(B_1,\dots,B_r)$ both generate the algebra $M_d(K)$. Then I claim that $(A_1,\dots,A_r)\stackrel\chi\sim(B_1,\dots,B_r)$ if and only if there is some invertible $X$ where $B_j=XA_jX^{-1}$ for $1\leq j\leq r$ (if $B_j=XA_jX^{-1}$ for $1\leq j\leq r$, then we say that $(A_1,\dots,A_r),(B_1,\dots,B_r)$ are jointly similar). The direction $\leftarrow$ is clear, so we just need to assume $(A_1,\dots,A_r)\stackrel\chi\sim(B_1,\dots,B_r)$ in order to establish the direction $\rightarrow$.
Suppose that $(A_1,\dots,A_r),(B_1,\dots,B_r)$ both generate the algebra $M_d(K)$. Then define an equivalence relation $\approx$ on $K\langle x_1,\dots,x_r\rangle$ where we set
$p_0(x_1,\dots,x_r)\approx p_1(x_1,\dots,x_r)$ precisely when
$$\chi((p_0(A_1,\dots,A_r)-p_1(A_1,\dots,A_r))q(A_1,\dots,A_r))=x^d$$ for each $q\in K\langle x_1,\dots,x_r\rangle$.
Then we observe that $p_0(x_1,\dots,x_r)\approx p_1(x_1,\dots,x_r)$ if and only if
$p_0(A_1,\dots,A_r)=p_1(A_1,\dots,A_r).$ Therefore, the mapping
$\phi_0:K\langle x_1,\dots,x_r\rangle/\approx\rightarrow M_d(K)$ defined by letting
$\phi(p(x_1,\dots,x_r)/\approx)=p(A_1,\dots,A_r)$ is a $K$-algebra isomorphism.
On the other hand, we can define an isomorphism $\phi_1:K\langle x_1,\dots,x_r\rangle/\simeq\rightarrow M_d(K)$ by letting
$\phi_1(p(x_1,\dots,x_r)/\approx)=p(B_1,\dots,B_r)$. We therefore obtain an automorphism $\phi_1\phi_0^{-1}:M_d(K)\rightarrow M_d(K)$ where
$$\phi_1\phi_0^{-1}(p(A_1,\dots,A_r))=\phi_1(p(x_1,\dots,x_r)/\approx)=p(B_1,\dots,B_r).$$ Since $\phi_1\phi_0^{-1}$ is an automorphism of the algebra $M_d(K)$, there is some invertible $X$ where
$\phi_1\phi_0^{-1}(Y)=XYX^{-1}$ for all $X$. Therefore, we have
$$p(B_1,\dots,B_r)=\phi_1\phi_0^{-1}(p(A_1,\dots,A_r))=X\cdot p(A_1,\dots,A_r)\cdot X^{-1}$$ for all $p$, so $(A_1,\dots,A_r)$ and $(B_1,\dots,B_r)$ are jointly similar.
I suspect that we can also prove our main result without relying on the fact that the only algebra automorphisms of $M_d(K)$ are inner, and perhaps I will come back to this answer with that other proof.
The hypothesis that $(A_1,\dots,A_r)$ generates $M_d(K)$ is a weak hypothesis that is satisfied in almost all cases. For example, if $A=(a_{i,j})$ and $B=(b_{i,j})$ are the matrices where $a_{i,j}=1$ if $j=i+1\mod d$ and $a_{i,j}=0$ otherwise and where $b_{i,j}=1$ iff $i=j=1$ and $b_{i,j}=0$ otherwise, then
$\\{A^rBA^s,0\leq r<d,0\leq s<d\\}$ is the canonical basis for the vector space $M_d(K)$. If $K$ is an infinite field, then the set of all pairs $(A,B)$ of $d\times d$-matrices where $\\{A^rBA^s,0\leq r<d,0\leq s<d\\}$ is a basis for the vector space $M_d(K)$ is a non-empty Zariski open set. If $V$ is an affine space over an infinite field $K$, and $O_i$ is Zariski open in $V$ for $i\in I$, and $|I|<|K|$, then $\bigcap_{i\in I}O_i\neq\emptyset$ (reference). This means that subsets of proper Zariski closed set should be considered to be 'small' sets. We conclude that if $r>1$ and the field $K$ is infinite, then $(A_1,\dots,A_r)$ typically generates the algebra $M_d(K)$.
I personally find it easier to work with non-commuting matrices $(A_1,\dots,A_r)$ when $(A_1,\dots,A_r)$ generates the algebra $M_d(K)$.
