This is a partial answer to the case where we aren't requiring that $W$ be closed under multiplication. Specifically, this is the case when $W$ has codimension $1$. I'm going to assume $V=F^n$, where $F$ is our base field, so $\mbox{End}(V)=M_n(F)$.

In this case, we can describe $W$ by a nonzero matrix $A$, i.e., $M\in W \iff A\cdot M=0$, where by $\cdot$ I mean we are actually "dotting" the matrices as if they were vectors, $A\cdot B=\sum_{i,j=1}^n a_{ij}b_{ij}$.

Claim: $W$ can fix any vector if and only if $A$ either has trace $0$ or has rank greater than $1$.

First, some observations: Obviously, scaling $A$ doesn't change $W$, so we can freely scale $A$. Furthermore, if we conjugate $A$, this doesn't change whether $W$ can fix any vector.

This is because for any matrix $S$, we have $SA\cdot M=A \cdot S^T M$, and $AS\cdot M=A \cdot MS^T$. The first is because because $A\cdot M$ is just the sum of the dot products of the columns of $A$ with the corresponding columns of $M$, and the second is because of the same fact with the rows. So if $W'$ is the space corresponding to $SAS^{-1}$, then $W'=S^T W (S^T)^{-1}$. So if $v\in V$ and $W$ can fix any vector, then $W$ can fix $(S^T)^{-1} v$, which means that $W'$ can fix $v$.

So we can freely scale and conjugate $A$. Assume for now that $F$ is algebraically closed, to make things easier; then we can assume that $A$ is in Jordan normal form.

Suppose $W$ fails to fix some vector. Then $A$ must have nonzero trace, as otherwise we would have $I\in W$. Since $A$ has nonzero trace, $A$ has some nonzero eigenvalue, and so $A$ has some row consisting of a nonzero element on the diagonal and zeroes elsewhere. (Really, this was all we needed algebraic closure for.) Conjugating and scaling, we'll assume this element is a $1$ and is in the lower right corner.

So now we have the question, for any $v=(x_1,\ldots,x_n)\in F^n$, does there exist $B=(b_{ij})_{i,j=1}^n\in M_n(F)$ such that $B$ fixes $v$ and $A\cdot B=0$? Well, for any given $v$, this is just a system of linear equations in the $b_{ij}$. Since we assumed $a_{jj}=1$, we'll exclude $b_{jj}$ as a parameter, writing it in terms of the others instead.

The $n \times n^2-1$ matrix for this system of linear equations (where the columns $b_{ij}$ are put in order first by row and then by column) is as follows: In row $i$, for $i\lt n$, the nonzero entries run from $n(i-1)+1$ to $ni$, being $x_1, \ldots, x_n$ in turn. In row $n$, we have $x_1,\ldots,x_{n-1}$ in positions $n(n-1)+1$ to $n^2-1$ as before, but of course there is no column $n^2$, and also there are scattered $-a_{ij}$ in the other columns. (These don't affect the final $x_1,\ldots,x_{n-1}$ because of the assumption that $a_{nj}=0$ for $j\lt n$.)

If any of $x_1,\ldots,x_{n-1}$ are nonzero, this matrix clearly has full rank, and so $W$ can fix $v$. Whereas if $x_1=\ldots=x_{n-1}=0$ and $x_n\ne 0$, then $W$ can clearly again fix $v$... except possibly if each one of the scattered $-a_{ij}$ is in the same column as some $x_n$ above, i.e., if the only nonzero entries in $A$ are in the last column. Since we assumed $W$ does not fix some vector, we conclude that $A$ has rank $1$.

Thus, if $W$ fails to fix some vector, $A$ has nonzero trace and rank $1$. For the converse, if $A$ has nonzero trace and rank $1$, then once again we can put $A$ in Jordan normal form. There are only two possible Jordan normal forms for a matrix of rank $1$ -- a lone nonzero element on the diagonal, and a lone $1$ just above the diagonal. Since $A$ has nonzero trace, we're in the former case. Scaling and conjugating again, we can assume $A$ is the matrix with a lone $1$ in the upper left corner, so $W$ is the space of all matrices with a $0$ in the upper left corner. But then $W$ cannot fix $(1,0,\ldots,0)$, so $W$ fails to fix some vector.

This concludes the proof when $F$ is algebraically closed. If $F$ is not algebraically closed, well, all the equations we're trying to solve are linear, so this makes no difference. More concretely, if $A$ has nonzero trace and rank $1$, it's still similar to the matrix with a lone $1$ in the upper left corner (similarity is invariant under field extension), and so $W$ still fails to fix some vector. Otherwise, for a given vector $v$, finding $B$ that fixes $v$ and satisfies $A\cdot B=0$ is still just solving a system of linear equations in the $b_{ij}$, so if there is a solution over $\overline{F}$, there is a solution over $F$ (e.g. take a solution over $\overline{F}$ and average it with its conjugates). This proves the claim.