Function "equivariant" to linear transforms Let $f\colon \mathbb{R}^{m\times n}\to  \mathbb{R}^{m \times n}$ be such that
$$f(AX+b{\bf 1}^T) = Af(X)$$
for any $X\in \mathbb{R}^{m\times n}$, $A\in \mathbb{R}^{m\times m}$, and $b\in\mathbb{R}^m$. The notation $b{\bf 1}^T$ means an outer product (broadcasting $b$ as columns).
I would like to show that $f$ must take the form $f(X) = X C$ for some $n\times n$ matrix $C$. Is it true? If so, what are implied constraints on $C$? Very appreciate your help.
[Edit] it seems that it is not true under these conditions. I will give a counter-example with $m=1$ a bit later. I realized my actual case of interest does not require it for any arbitrary $X$ but only for one given $X$. In this case $C$ can in fact depend on $X$. So the question reformulates as: does there always exist representation as $f(X)=XC(X)$?
 A: (The following answer addresses your original question, in which the $X$ in the condition is not fixed but arbitrary.) It is true when $m>1$. Let $\{e_1,\ldots,e_m\}$ be the standard basis of $\mathbb R^m$. By putting $A=e_ie_j^T$ and $b=0$ into the given condition, we have
$$
e_i^Tf(e_ie_j^TX)=e_i^Te_ie_j^Tf(X)=e_j^Tf(X)\tag{1}
$$
for all $i,j\in\{1,2,\ldots,m\}$ and all matrices $X$. Define $g_i:\mathbb R^{1\times n}\to\mathbb R^{1\times n}$ by $g_i(x^T)=e_i^Tf(e_ix^T)$. Then $(1)$ can be rephrased as
$$
g_i(e_j^TX) =e_j^Tf(X).\tag{2}
$$
Hence $g_i(x^T)=g_i(e_j^Te_jx^T)=e_j^Tf(e_jx^T)$ for all indices $i,j$ and all $x^T\in\mathbb R^{1\times n}$. Consequently, all $g_i$s are identical to some common function $g$. Now, for any $x^T,y^T\in\mathbb R^{1\times n}$,
\begin{aligned}
g(x^T+y^T)
&=g((e_1+e_2)^T(e_1x^T+e_2y^T))\\
&=g\left(e_1^Te_1(e_1+e_2)^T(e_1x^T+e_2y^T)\right)\\
&=e_1^Tf\left(e_1(e_1+e_2)^T(e_1x^T+e_2y^T)\right)\\
&=e_1^Te_1(e_1+e_2)^Tf(e_1x^T+e_2y^T)\\
&=e_1^Tf(e_1x^T+e_2y^T)+e_2^Tf(e_1x^T+e_2y^T)\\
&=g\left(e_1^T(e_1x^T+e_2y^T)\right)+g\left(e_2^T(e_1x^T+e_2y^T)\right)\\
&=g(x^T)+g(y^T).\\
\end{aligned}
(Note that $e_2$ exists in the above because $m>1$.) Hence $g$ is additive. It also preserves scalar multiplication, because
\begin{aligned}
g(ax^T)
&=g\left(e_1^T(ae_1x^T)\right)\\
&=e_1^Tf(ae_1x^T)\\
&=e_1^Tf\left((aI_m)e_1x^T\right)\\
&=e_1^T(aI_m)f(e_1x^T)\\
&=ae_1^Tf(e_1x^T)\\
&=ag(x^T).
\end{aligned}
Therefore $g$ is linear. Hence $g(x^T)=x^TC$ for some matrix $C\in\mathbb R^{n\times n}$. It follows from $(2)$ that
$$
f(X)=\pmatrix{e_1^Tf(X)\\ e_2^Tf(X)\\ \vdots\\ e_m^Tf(X)}
=\pmatrix{g(e_1^TX)\\ g(e_2^TX)\\ \vdots\\ g(e_m^TX)}
=\pmatrix{e_1^TXC\\ e_2^TXC\\ \vdots\\ e_m^TXC}
=XC.
$$
Note that we only need $f(AX)=Af(X)$ to conclude that $f(X)=XC$ and the same argument holds when $\mathbb R$ is replaced by any commutative ring with unity (in other words, when $m>1$ and $R$ is a commutative ring, the only functions on $R^{m\times n}$ that commute with ALL left matrix multiplications are right matrix multiplications). However, the full condition $f(AX+b\mathbf1^T)=Af(X)$ implies that $\mathbf1^TC$ must be zero.
The case where $m=1$ is more anarchic. When $n=1$, $f$ is necessarily zero. When $n=2$, we have $f(b+a,b-a)=af(1,-1)$ and hence $f$ is linear. When $n\ge3$, $f$ may take a more arbitrary form. In fact, for each unit vector $u\perp\mathbf1$ whose first nonzero element is positive, one may assign an arbitrary value to $f(u^T)$ and define $f(au^T+b\mathbf1^T)=af(u^T)$. Then $f$ preserves scalar multiplication but it is non-linear in general.
