Is this piecewise-defined system time-invariant?
Let us suppose that the system with input $x$ is defined by:
\begin{equation}
  y[n] =
    \begin{cases}
      x[n] &amp; \text{if} &amp; n&lt;0\\
      2x[n] &amp; \text{if} &amp; n\ge 0
    \end{cases}       
\end{equation}
As answered by @TimWescott, the two-part  definition suggests a lack of time invariance. Finding a counter-example could be a good start. For instance a two-part signal whose behavior when multiplied by 2 is visible. Indeed a modification of the Heavide jump is interesting:
\begin{equation}
  x_0[n] =
    \begin{cases}
      -1&amp; \text{if} &amp; n&lt;0\\
      \phantom{-}1 &amp; \text{if} &amp; n\ge 0
    \end{cases}       
\end{equation}
along with the right-shifted version:
\begin{equation}
  x_1[n] =
    \begin{cases}
      -1 &amp; \text{if} &amp; n&lt;1\\
      \phantom{-}1 &amp; \text{if} &amp; n\ge 1\,.
    \end{cases}       
\end{equation}
You get:
\begin{equation}
  y_0[n] =
    \begin{cases}
      -1&amp; \text{if} &amp; n&lt;0\\
      \phantom{-}2 &amp; \text{if} &amp; n\ge 0
    \end{cases}       
\end{equation}
along with the output for the right-shifted version:
\begin{equation}
  y_1[n] =
    \begin{cases}
      -1 &amp; \text{if} &amp; n&lt;0\\
      -2 &amp; \text{if} &amp; n=0\\
      \phantom{-}2 &amp; \text{if} &amp; n\ge 1\,.
    \end{cases}       
\end{equation}
which are not time-shifted versions of each other.