At least when $n=1$, there are no nontrivial transformations of this kind.
Indeed, suppose that $v(t,x)=u(\tau(t,x),\xi(t,x))$, where $u_t=u_{xx}$. Then $u_{tx}=u_{xxx}$ and $u_{tt}=u_{xxxx}$, so that the equation\begin{equation*} v_t=v_{xx} \tag{1}\label{1} \end{equation*}can be rewritten as\begin{equation*} \xi_x(u_{xxx}\tau_x+u_{xx}\xi_x)+\tau_x\,(u_{xxxx}\tau_x+u_{xxx}\xi_x) \\ +u_{xx}\tau_{xx}+u_x\xi_{xx}-u_{xx}\tau_t-u_x\xi_t=0. \tag{2}\label{2} \end{equation*}Since at any point $(t,x)$ the partial derivatives $u_x,u_{xx},u_{xx},u_{xxx},u_{xxxx}$ can take arbitrary real values, the coefficients of $u_x,u_{xx},u_{xx},u_{xxx},u_{xxxx}$ in \eqref{2} must all be zero. In particular, the coefficient $\tau_x^2$ of $u_{xxxx}$ in \eqref{2} must be $0$, so that $\tau_x=\tau_{xx}=0$, and we can write $\tau(t,x)=T(t)$ for some function $T$.
So, \eqref{2} implies\begin{equation*} u_{xx}(\xi_x^2-T'(t))+u_x(\xi_{xx}-\xi_t)=0,\end{equation*}so that\begin{equation*} \xi_x^2=T'(t),\quad\xi_{xx}=\xi_t. \end{equation*}Hence, $\xi_x\xi_{xx}=0$ and, if $\xi$ is real analytic, then\begin{equation*}\xi_{xx}=0, \quad\text{whence}\quad\xi_t=0, \end{equation*}whence for some real $a,b,c$ we have $\xi(t,x)=a+bx$, $T'(t)=\xi_x^2=b^2$, $\tau(t,x)=T(t)=c+b^2t$. Thus,\begin{equation*} \xi(t,x)=a+bx,\quad\tau(t,x)=c+b^2t.\end{equation*}
The case $n\ge2$ should be similar.
