If $ u $ satisfy the heat equation $ u_{t} = ku_{xx} $ then so does $ u_{\alpha, \beta, \gamma} $ ( where $ u_{\alpha, \beta, \gamma}(x,t) = \gamma u(\alpha x, \beta t) $), provided $ \beta = \alpha^{2} $.
My attempt:
$ \dfrac{\partial}{\partial t} u_{\alpha,\beta,\gamma}(x,t) = \gamma\beta\dfrac{du}{dt}$
$ \dfrac{\partial}{\partial x^{2}} u_{\alpha,\beta,\gamma}(x,t) = \gamma\beta\dfrac{d^{2}u}{dx^{2}}$
And here I'm stuck.
Note that $$ \partial_{t}u_{\alpha,\beta,\gamma}=\partial_t \gamma u(\alpha x,\beta t)\\ =\gamma \beta u_t(\alpha x,\beta t) $$ and $$ \partial_{xx}u_{\alpha,\beta,\gamma}=\partial_{xx} \gamma u(\alpha x,\beta t)\\ =\gamma \alpha^2 u_{xx}(\alpha x,\beta t) $$ so $$ \partial_{t}u_{\alpha,\beta,\gamma}-k\partial_{xx}u_{\alpha,\beta,\gamma}= \gamma \beta u_t(\alpha x,\beta t)-k\gamma \alpha^2 u_{xx}(\alpha x,\beta t)\\ =\gamma\left( \beta u_t(\alpha x,\beta t)-k \alpha^2 u_{xx}(\alpha x,\beta t) \right)\\ \stackrel{\alpha^2=\beta}{=} \gamma\beta\left( u_t(\alpha x,\beta t)-k u_{xx}(\alpha x,\beta t) \right)\\ =0 $$ since $u$ solves the heat equation.