Can $n$ extendible cardinals have trivial algebraic structure?

If $j$ is a function with $V_{\lambda+1}\subseteq\mathrm{Dom}(f)$, then define a mapping $j\upharpoonright_{\lambda+1}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ by letting $j\upharpoonright_{\lambda+1}(A)=j(A)\cap V_{\lambda}$ for each $A\subseteq V_{\lambda}$.

Suppose that $\lambda$ is an inaccessible cardinal. Then we shall say that a function $f:V_{\lambda+1}\rightarrow V_{\lambda+1}$ is an extendibility mapping if for all $\alpha>\lambda$ there is some elementary embedding $j:V_{\alpha}\rightarrow V_{\beta}$ where $f=j\upharpoonright_{\lambda+1}$. If $\kappa<\lambda$ and $\lambda$ is inaccessible, then $\kappa$ is extendible if and only if there is some extendibility mapping $f:V_{\lambda+1}\rightarrow V_{\lambda+1}$ with $\mathrm{crit}(f)=\kappa.$

Let $\Phi(f_{1},...,f_{n},\lambda)$ denote the following formula.

i. $f_{1},...,f_{n}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ are non-trivial extendibility mappings with $\mathrm{crit}(f_{1})<...<\mathrm{crit}(f_{n})<\lambda$ and $\lambda$ inaccessible, and

ii. whenever $1\leq i<j\leq n$ and $\alpha>\lambda$ and $j:V_{\alpha}\rightarrow V_{\beta}$ is an elementary embedding with $j\upharpoonright_{\lambda+1}=f_{j}$ then $(j(f_{i}))\upharpoonright_{\lambda+1}=f_{i}$.

Is the existence of $n$ extendible cardinals equiconsistent with the existence of extendibility mappings $f_{1},...,f_{n}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ where $\Phi(f_{1},...,f_{n},\lambda)$ is false?

The motivation behind asking this question is that the extendible cardinals begin to have some aspects of the algebras of elementary embeddings and these algebraic properties of the extendible cardinals may be easier to investigate than the algebraic properties at the level around $n$-hugeness and rank-into-rank. I will probably ask more questions like this one in the near future.