Let $X\subset\mathbb{A}^n$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristic zero. Suppose we have two linearly reductive algebraic group $G, G'$ such that there's a closed embedding $\rho:G\hookrightarrow G'$ and there's a regular action of $G$ on $X$ and a regular action of $G'$ on $\mathbb{A}^m$, $m>n$.

Is there a "natural" way to embed $\Phi:X\rightarrow\mathbb{A}^m$ such that $\Phi$ is equivariant, i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in G$.

My setting: Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.

For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{K}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis. Now take an irreducible component $X\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $X$.

Corresponding to $X$ we have an affine space $\rep_{\alpha}(Q):=\prod_{a\in Q_1}\operatorname{Mat}_{\alpha(ha)\times\alpha(ta)}(\mathbb{K})$, and $\rep_{\alpha}(Q)$ has an open set $U$ such that for any representation $V\in U$, we have $V/IV\cong W$, where $W\in X$ is general representation. Similarly, $\rep_{\alpha}(Q)$ has the action of $\GL_{\alpha}$ (defined similarly as $\GL_{\beta}$) by change of basis. Similarly we can define $\rep_{\beta}(Q)$.

This dimension vector $\alpha$ is such that $\alpha(i)\geq\beta(i)$, for every $i\in Q_0$. We get an embedding of $\rho:\GL_{\beta}\hookrightarrow\GL_{\alpha}$ via $(g_i,i\in Q_0)\mapsto(h_i,i\in Q_0)$, where $h_i$ is obtained by adjoining identity matrix of appropriate dimension in the diagonal of $g_i$.

So my question becomes:

Is there a "natural" way to get a closed embedding $\Phi:X\hookrightarrow\rep_{\beta}(Q)\rightarrow\rep_{\alpha}(Q)$ such that $\Phi$ is equivariant i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in\GL_{\beta}$.

Let $X\subset\mathbb{A}^n$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristic zero. Suppose we have two linearly reductive algebraic groups $G$, $G'$ such that there's a closed embedding $\rho:G\hookrightarrow G'$ and there's a regular action of $G$ on $X$ and a regular action of $G'$ on $\mathbb{A}^m$, $m>n$.

Is there a "natural" way to embed $\Phi:X\rightarrow\mathbb{A}^m$ such that $\Phi$ is equivariant, i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in G$?

My setting: Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.

For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{K}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis. Now take an irreducible component $X\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $X$.

Corresponding to $X$ we have an affine space $\rep_{\alpha}(Q):=\prod_{a\in Q_1}\operatorname{Mat}_{\alpha(ha)\times\alpha(ta)}(\mathbb{K})$, and $\rep_{\alpha}(Q)$ has an open set $U$ such that for any representation $V\in U$, we have $V/IV\cong W$, where $W\in X$ is general representation. Similarly, $\rep_{\alpha}(Q)$ has the action of $\GL_{\alpha}$ (defined similarly as $\GL_{\beta}$) by change of basis. Similarly we can define $\rep_{\beta}(Q)$.

This dimension vector $\alpha$ is such that $\alpha(i)\geq\beta(i)$, for every $i\in Q_0$. We get an embedding of $\rho:\GL_{\beta}\hookrightarrow\GL_{\alpha}$ via $(g_i,i\in Q_0)\mapsto(h_i,i\in Q_0)$, where $h_i$ is obtained by adjoining identity matrix of appropriate dimension in the diagonal of $g_i$.

So my question becomes:

Is there a "natural" way to get a closed embedding $\Phi:X\hookrightarrow\rep_{\beta}(Q)\rightarrow\rep_{\alpha}(Q)$ such that $\Phi$ is equivariant, i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in\GL_{\beta}$?

Let $X\subset\mathbb{A}^n$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristic zero. Suppose we have two linearly reductive algebraic group $G, G'$ such that there's a closed embedding $\rho:G\hookrightarrow G'$ and there's a regular action of $G$ on $X$ and a regular action of $G'$ on $\mathbb{A}^m$, $m>n$.

Is there a "natural" way to embed $\Phi:X\rightarrow\mathbb{A}^m$ such that $\Phi$ is equivariant, i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in G$.

My setting: Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.

For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{K}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis. Now take an irreducible component $X\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $X$.

Corresponding to $X$ we have an affine space $\rep_{\alpha}(Q):=\prod_{a\in Q_1}\operatorname{Mat}_{\alpha(ha)\times\alpha(ta)}(\mathbb{K})$, and $\rep_{\alpha}(Q)$ has an open set $U$ such that for any representation $V\in U$, we have $V/IV\cong W$, where $W\in X$ is general representation. Similarly, $rep_{\alpha}(Q)$ has the action of $\GL_{\alpha}$ (defined similarly as $\GL_{\beta}$) by change of basis. Similarly we can define $\rep_{\beta}(Q)$.

This dimension vector $\alpha$ is such that $\alpha(i)\geq\beta(i)$, for every $i\in Q_0$. We get an embedding of $\rho:\GL_{\beta}\hookrightarrow\GL_{\alpha}$ via $(g_i,i\in Q_0)\mapsto(h_i,i\in Q_0)$, where $h_i$ is obtained by adjoining identity matrix of appropriate dimension in the diagonal of $g_i$.

So my question becomes:

Is there a "natural" way to get a closed embedding $\Phi:X\hookrightarrow\rep_{\beta}(Q)\rightarrow\rep_{\alpha}(Q)$ such that $\Phi$ is equivariant i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in\GL_{\beta}$.

Let $X\subset\mathbb{A}^n$ be an irreducible affine variety over an algebraically closed field $\mathbb{K}$ of characteristic zero. Suppose we have two linearly reductive algebraic group $G, G'$ such that there's a closed embedding $\rho:G\hookrightarrow G'$ and there's a regular action of $G$ on $X$ and a regular action of $G'$ on $\mathbb{A}^m$, $m>n$.

Is there a "natural" way to embed $\Phi:X\rightarrow\mathbb{A}^m$ such that $\Phi$ is equivariant, i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in G$.

My setting: Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.

For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{K}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis. Now take an irreducible component $X\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $X$.

Corresponding to $X$ we have an affine space $\rep_{\alpha}(Q):=\prod_{a\in Q_1}\operatorname{Mat}_{\alpha(ha)\times\alpha(ta)}(\mathbb{K})$, and $\rep_{\alpha}(Q)$ has an open set $U$ such that for any representation $V\in U$, we have $V/IV\cong W$, where $W\in X$ is general representation. Similarly, $\rep_{\alpha}(Q)$ has the action of $\GL_{\alpha}$ (defined similarly as $\GL_{\beta}$) by change of basis. Similarly we can define $\rep_{\beta}(Q)$.

This dimension vector $\alpha$ is such that $\alpha(i)\geq\beta(i)$, for every $i\in Q_0$. We get an embedding of $\rho:\GL_{\beta}\hookrightarrow\GL_{\alpha}$ via $(g_i,i\in Q_0)\mapsto(h_i,i\in Q_0)$, where $h_i$ is obtained by adjoining identity matrix of appropriate dimension in the diagonal of $g_i$.

So my question becomes:

Is there a "natural" way to get a closed embedding $\Phi:X\hookrightarrow\rep_{\beta}(Q)\rightarrow\rep_{\alpha}(Q)$ such that $\Phi$ is equivariant i.e., $\Phi(g\cdot x)=\rho(g)\cdot \Phi(x)$, for any $x\in X$ and any $g\in\GL_{\beta}$.