Distinguishing combinatorial maps by their linearizations Every (not-necessarily invertible) map $f$ from $[n]:=\{1,2,,,,.n\}$ to itself determines a linear map $L_f$ from ${\bf R}^n$ to itself that sends the basis vector $e_k$ to $e_{f(k)}$ for $1 \leq k \leq n$. 
If $f$ and $g$ are two self-maps of $[n]$ for which the associated linear maps $L_f$ and $L_g$ are conjugate in the sense that there exists $A$ in $GL(n)$ with $L_f \circ A = A \circ L_g$, must $f$ and $g$ be conjugate in the sense that there exists a permutation $\pi$ in $S(n)$ with $f \circ \pi = \pi \circ g$? 
Note that if we restrict to the case where $f$ and $g$ are permutations, the answer is Yes.
In the unrestricted case, I'm fairly sure the answer is No; Jordan canonical form (IIRC) classifies linear maps up to conjugation by $GL(n)$, and the decomposition into Jordan blocks is too combinatorially simple to capture the kind of tree structure self-maps of sets can possess. On the other hand, I don't see how to turn this intuition into a counterexample.
In the event that the answer is No (as I expect), then I ask a follow-up question: Is there some more refined way to linearize $f$ so that the resulting map, considered up to linear-map conjugacy, captures all the combinatorics of $f$?
 A: As Benjamin Steinberg comments, the problem reduces to the following:
represent $f$ by the digraph with vertices $[n]$ and edges $i\to
f(i)$. Delete all the cycles. What remains is an acyclic digraph $D$
that defines a nilpotent matrix $N$ whose rows and columns are indexed
by the vertices of $D$. Namely, $N_{uv}=1$ if there is an edge $u\to
v$, and $N_{uv}=0$ otherwise. We need to find the Jordan form of $N$.
Now more generally let $E$ be any finite acyclic digraph and for each
edge $u\to v$ let $x_{uv}$ be an indeterminate. Define $N_{uv}=x_{uv}$
if $u\to v$, and $N_{uv}=0$ otherwise. The Jordan form of this
"generic nilpotent matrix" was determined by M. Saks and E. R. Gansner
(independently). For Gansner's paper see
http://www2.research.att.com/~erg/pdf/JADM-Gansner81.pdf.  Namely, the
sum of the sizes of the largest $j$ Jordan blocks is equal to the
largest number of vertices in a union of $j$ paths of $E$. If we
replace $x_{uv}$ by 1, then the problem is no longer
combinatorial. For instance, let $A$ be any $n\times n$
$(0,1)$-matrix, and place it in the upper-right corner of a
$(2n)\times (2n)$ matrix $B$ whose other entries are 0. Then $A$ and
$B$ have the same rank so the Jordan form of the nilpotent matrix $B$
depends on the rank of an arbitrary $(0,1)$-matrix $A$, for which
there is not a purely combinatorial description. However, going back
to $D$, because $N$ has at most one 1 in each row it is easy to see
that the Gansner-Saks theorem will continue to hold for the matrix
$N$. That is, the sum of the sizes of the largest $j$ Jordan blocks is
equal to the largest number of vertices in a union of $j$ paths of
$D$.
A: The matrices
$A = \left(
\begin{matrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1
\end{matrix}
\right)$
and
$B = \left(
\begin{matrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1
\end{matrix}
\right)$
are similar (in fact, $A = XBX^{-1}$ for
$X = \left(
\begin{matrix}
0 & -1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{matrix}
\right)$), but are of the form $L_f$ for two non-isomorphic $f$'s, right?
I am wondering how the Jordan normal form of $L_f$ can be derived combinatorially from the structure of $f$.
A: Updated answer.
Here is a more complete answer to your question including the follow-up question. Define the rank of a mapping $f$ to be the size of its image. 
Thm. TFAE.


*

*$L_f$ and $L_g$ are conjugate. 

*The images of $f$ and $g$ are conjugate under any linear representation of the monoid of all mappings over any field.  

*The rank of $f^j$ is the same as the rank of $g^j$ for all $1\leq j\leq n$ and the number of cycles of $f$ of size $i$ equals the number of cycles of $g$ of size $i$ for all $1\leq i\leq n$.  


The point for the equivalence of 1 and 3 is that the number of Jordan blocks of size $\geq j$ in the nilpotent block of the Fitting decomposition of $L_f$ is the rank of $f^j$ minus the rank of $f^{j+1}$ and the invertible part of the Fitting decomposition is the permutation matrix associated to its action on the elements in its cycles. 
This extends to any representation by noting that if $f$ and $g$ are mappings of the same rank then $f=xgy$ and $h=ugv$ for some maps $x,y,u,v$ and so they map to matrices of the same rank under any representation over any field. Then the above argument essentially runs through for any representation.  
Of course the rank condition can be expressed in digraph language in the form discussed by Richard Stanley but I found the above formulation useful when trying to answer a similar question for arbitrary finite semigroups. 
Original answer. Here is a more egregious example which shows some maps go to linearly conjugate matrices under any representation but are not conjugate under the symmetric group. 
Two idempotent mappings will give rise to linearly conjugate matrices under your representation  if and only if they have the same rank (same cardinality images).  But to be conjugate under $S_n$ the integer partition of n obtained from counting preimages of each element of the image would have to be the same.
Idempotent mappings  of the same rank will in fact go to idempotent matrices of the same rank under any representation of the full transformation monoid because they generate the same two sided ideal in the monoid.  Since idempotent matrices are conjugate if and only if they have the same rank this gives no to your second question.
So for example the idempotent which maps 1 to 1 and all other elements to 2 is not conjugate under $S_n$ to an idempotent that maps the even numbers to 2 and the odd ones to 1 (for $n>3$) but both are linearly conjugate under any representation.
