Q:

Can I say that if a kernel is translation-invariant (or shift-invariant), then the kernel is symmetric?

Let $x,y \in \mathbb{R}$ where $y=x-t$. Translation-invariant (or shift-invariant) kernel $\kappa(\cdot,\cdot)$ is defined as
$\kappa(x,y) = \kappa(x,x-t) = \kappa(t)$. 
Can I say that the function $\kappa$ is symmetric ?
I think "yes", if I can define the translation-invariance as "DIFFERENCE between $x$ and $y$" so that $y=x+t$. And $\kappa(x,y)=\kappa(x,x+t)=\kappa(-t)$.
But, wanted make sure the relation between translation-invariance and symmetry (i.e., $\kappa(x,y)=\kappa(y,x)$).
Can I simply say that translation invariant kernel is $\kappa(x,y) = \kappa(x-y) = \kappa(y-x)$ ?
translation-invariant $\to$ symmetric. Is this always correct???
Or, is the symmetry required for the second equality?

A:

Unfortunately I still don't understand the parts of your question where you try to motivate the idea that translation invariance might imply symmetry, so I can only state that this implication doesn't hold. If a kernel $\kappa(x,y)$ is translation-invariant and is thus given by $\kappa(x,y)=g(x-y)$ for some function $g$, then $\kappa$ is symmetric in the sense $\kappa(x,y)=\kappa(y,x)$ if and only if $g$ is symmetric in the sense $g(t)=g(-t)$, since $\kappa(x,y)=g(x-y)$ and $\kappa(y,x)=g(y-x)=g(-(x-y))$. For instance, for symmetric $g(t)=t^2$, the kernel is symmetric with $\kappa(x,y)=\kappa(y,x)=(x-y)^2$, while for antisymmetric $g(t)=t$, the kernel is antisymmetric with $\kappa(x,y)=-\kappa(y,x)=x-y$.

