A question on 'equivariant' splittings of a short exact sequence of $K$-vector spaces Let $K$ be a field and
$$
\begin{array}{rcccccccl}
0 &\rightarrow & V' & \xrightarrow{a} & V & \xrightarrow{b} & V'' & \rightarrow & 0\\
\end{array}
$$
be a short exact sequence of $K$-vector spaces. Each such sequence splits, i.e. there is a $K$-linear map $b':V''\to V$ such that $b\circ b'=id_{V''}$. 
Now suppose that we have given $K$-linear maps $f':V'\to V'$ and $f:V\to V$ and $f'':V''\to V''$ fitting (as the unlabeled vertical maps) into a diagram
$$
\begin{array}{rcccccccl}
0 &\rightarrow & V' & \xrightarrow{a} & V & \xrightarrow{b} & V'' & \rightarrow & 0\\
 & &\downarrow &&\downarrow & &\downarrow &\\
0 &\rightarrow & V' & \xrightarrow{a} & V & \xrightarrow{b} & V'' & \rightarrow & 0.\\
\end{array}
$$
Suppose that both $f'$ and $f''$ are diagonalizable with different eigenvalues, i.e. none of the eigenvalues of $f'$ is an eigenvalue of $f''$ and vice versa. Does one find an 'equivariant splitting' in this case, i.e. a $K$-linear map $b':V''\to V$ such that $b\circ b'=id_{V''}$ and $f\circ b'=b'\circ f''$?
 A: A pair $(V,f)$ with $V$ a vector space and $f:V\to V$ a linear map is the same thing as a $k[X]$-module with underlying vector space $V$. The map $f$ is diagonalizable iff the module is semisimple, and the eigenvalues are in that case correspond to the simple direct summands that $V$ has.
Using this, your question can be rephrased as:

if $0\to V'\to V\to V''\to0$ is a short exact sequence of $k[X]$ modules such that $V'$ and $V''$ are semisimple and do not have isomorphic direct summands, does the sequence split?

which in turn is the same as

if $V'$ and $V''$ are semisimple $k[X]$-modules with no isomorphic simple summands, is $Ext^1(V'',V')=0$?

and since $Ext$ is an additive functor, this is true if the answer to the following simpler question is yes:

if $V'$ and $V''$ are $1$-dimensional $k[X]$-modules and not isomorphic, is $Ext^1(V'',V')=0$?

The answer to this is indeed yes, and one can check it by hand. For example, you can translate the problem back to your original formulation, where it becomes:

if in the situation of your original question $V'$ and $V''$ are $1$-dimensional, does the sequence split equivariantly?

and this is very easy to do directly.

A little more work on the homological algebra side will prove thatin fact what you want is true even if $f'$ and $f''$ are not diagonalizable but merely do not have common eigenvalues, provided that $K$ is algebraically closed)
