**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.